LIBRARY OF CONGRESS. 



UNITED STATES OF AMERICA. 



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ELEMENTS OF GRAPHIC STATICS. 



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.m:^ 



THE 



ELEMENTS OF GRAPHIC STATICS 



a 2Ce.rt^Book for Stutirnts of lEngineering 



"-/ 



L. M. HOSKINS 

Professor of Pure and Applied Mechanics in the Leland 

Stanford Junior University; formerly Professor 

OF Mechanics in the University of Wisconsin 




NOV 12 1892 



MAC MIL LAN AND CO. 

AND LONDON 
1892 



All rights reserved 



K- 






Copyright, 1892, 
Bv MACMILLAN AND CO. 




Typography by J. S. Cushing & Co., Boston, U.S.A. 
Presswork by Berwick & Smith, Boston, U.S.A. 



^_ v;?6'?r 



PREFACE. 



The present work is designed as an elementary text-book for the 
use of students of engineering. In preparing it, a chief aim has been 
simpHcity of presentation. The matter treated has been Hmited to 
the development of fundamental principles, and their application fo the 
solution of typical problems. The method of the force and funicular 
polygons is deduced purely from statical principles, with very little 
consideration of the geometrical theory of reciprocal figures. Since 
the book is designed to embrace only what can profitably be taken in 
an elementary course by the student of engineering, it has not been 
thought best to include a discussion of problems involving the theory 
of elasticity. For similar reasons, the discussion of curves of inertia 
has been limited to simple cases ; a more general treatment being of 
interest to few besides the student of pure mathematics. No effort 
has been made to secure novelty in the matter treated, but it is 
believed that in a few cases it has been found possible to simpHfy, 
and perhaps thereby improve, the methods usually adopted. 

Attention is invited to the method adopted for lettering corre- 
sponding lines in force and space diagrams. It* will be seen that this 
is merely an extension of Bow's well-known notation. This notation 
is, however, capable of a much wider use than has usually been given 
it. It is believed that its use, wherever applicable, will be found of 
great value, both in facilitating the work of the student, and in 
guarding the draughtsman against mistakes. 

There is an unfortunate diversity of usage among writers in regard 
to the technical terms of mechanics, — a diversity especially notice- 
able in engineering Uterature. In this book the endeavor has been 
made in all cases to comply with the usage to which the highest 

authorities are tending. 

L. M. H. 

Madison, Wis., July, 1892. 



CONTENTS. 



PART I. — GENERAL THEORY. 
Chapter I. Definitions. — Concurrent Forces. 

PAGE 

§ I. Preliminary Definitions , i 

2. Composition of Concurrent Forces 5 

3. Equilibrium of Concurrent Forces 6 

4. Resolution of Concurrent Forces 8 

Chapter II Non-concurrent Forces. 

§ I. Composition of Ncn-concurrent Forces Acting on the Same Rigid 

Body II 

2. Equilibrium of Non-concurrent Forces 18 

3. Resolution into Non-concurrent Systems 28 

4. Moments of Forces and of Couples 30 

5. Graphic Determination of Moments 35 

6. Summary of Conditions of Equilibrium 37 

Chapter III. Internal Forces and Stresses. 

§ I. External and Internal Forces 40 

2. External and Internal Stresses 42 

3. Determination of Internal Stresses 46 

PART II. — STRESSES IN SIMPLE STRUCTURES. 

Chapter IV. Introductory. 

§ I . Outline of Principles and Methods 53 



viii CONTENTS. 

Chapter V. Roof Trusses. — Framed Structures Sustaining 

Stationary Loads. 

PAGE 

§ I. Loads on Roof Trusses 58 

2. Roof Truss with Vertical Loads 61 

3. Stresses Due to Wind Pressure 67 

4. Maximum Stresses 71 

5. Cases Apparently Indeterminate 74 

6. Three-hinged Arch 80 

7. Counterbracing 88 

Chapter VI. Simple Beams. 

§ I. General Principles 94 

2. Beam Sustaining Fixed Loads 98 

3. Beam Sustaining Moving Loads loi 



Chapter VII. Trusses Sustaining Moving Loads. 

Bridge Loads 112 

Truss Regarded as a Beam 115 

Truss with Parallel Chords Sustaining Concentrated Loads . . 117 

Parallel Chords — Uniformly Distributed Moving Load . . . . 130 

Truss with Curved Chords — Uniform Panel Loads 132 

Truss with Curved Chords — Concentrated Loads 137 



PART III.— CENTROIDS AND MOMENTS OF INERTIA. 
Chapter VIII. Centroids. 

§ I. Centroid of Parallel Forces 144 

2. Center of Gravity — Definitions and General Principles . . . . 149 

3. Centroids of Lines and of Areas 152 

Chapter IX. Moments of Inertia. 

§ J. Moments of Inertia of Forces 159 

2. Moments of Inertia of Plane Areas 169 

Chapter X. Curves of Inertia. 

§ I. General Principles 179 

2. Inertia-Ellipses for Systems of Forces 182 

3. Inertia-Curves for Plane Areas 187 



GRAPHIC STATICS. 



aXKc 



Part I. 
GENERAL THEORY, 



CHAPTER I. — DEFINITIONS. CONCURRENT 

FORCES. 

§ I. Preliminary Definitions. 

1. Dynamics treats of the action of forces upon bodies. Its 
two main branches are Statics and Kinetics. 

Statics treats of the action of forces under such conditions 
that no change of motion is produced in the bodies acted upon. 

Kinetics treats of the laws governing the production of motion 
by forces. 

2. Graphic Statics has for its object the deduction of the 
principles of statics, and the solution of its problems, by means 
of geometrical figures. 

3. A Force is that which tends to change the state of motion 
of a body. We conceive of a force as a push or a pull applied 
to a body at a definite point and in a definite direction. Such 
a push or pull tends to give motion to the body, but this ten- 
dency may be neutralized by the action of other forces. The 
effect of a force is completely determined when three things 
are given, — its magnitude, its direction, and its poi7tt of applica- 
tion. The line parallel to the direction of the force and con- 
taining its point of application, is called its line of action. 



2 GRAPHIC STATICS. 

Every force acting upon a body is exerted by some other body. 
But the problems of statics usually concern only the body acted 
upon. Hence, frequently, no reference is made to the bodies 
exerting the forces. 

4. Unit Force. — The tinit force is a force of arbitrarily chosen 
magnitude, in terms of which forces are expressed. Several 
different units are in use. The one employed in this work is 
thQ pound, which will now be defined. 

A pound force is a force equal to the weight of a pound mass 
at the earth's surface. A pound mass is the quantity of matter 
contained in a certain piece of platinum, arbitrarily chosen, and 
established as the standard by act of the British Parliament. 

The pound force, as thus defined, is not perfectly definite, 
since the weight of any given mass (that is, the attraction of 
the earth upon it) is not the same for all positions on the earth's 
surface. The variations are, however, unimportant for most of 
the requirements of the engineer. 

In its fundamental meaning, the word "pound" refers to the 
unit mass, and it is unfortunate that it is also applied to the unit 
force. The usage is, however, so firmly established that it will 
be here followed. 

5. Concurrent and Non-concurrent Forces. — Forces acting on 
the same body are concurrent when they have the same point of 
application. When applied at different points they are non- 
concurrent. 

6. Complanar Forces are those whose lines of action are in 
the same plane. In this work, only complanar systems will be 
considered unless otherwise specified. 

7. A Couple is the name given to a system consisting of two 
forces, equal in magnitude, but opposite in direction, and having 
different lines of action. The perpendicular distance between 
the two lines of action is called the arm of the couple. 



PRELIMINARY DEFINITIONS. 3 

8. Equivalent Systems of Forces. —Two systems of forces 
are equivalent when either may be substituted for the other 
without change of effect. 

9. Resultant. — A single force that is equivalent to a given 
system of forces is called the resultant of that system. It will 
be shown subsequently that a system of forces may not be 
equivalent to any single force. When such is the case, the 
simplest system equivalent to the given system may be called 
its resultant. Any forces having a given force for their 
resultant are called components of that force. 

10. Composition and Resolution of Forces. — Having given 
any system of forces, the process of finding an equivalent system 
is called the composition of fo7xes^ if the system determined con- 
tains fewer forces than the given system. If the reverse is the 
case, the process is called the resolution of forces. 

The process of finding the resultant of any given forces is 
the most important case of composition; while the process of 
finding two or more forces, which together are equivalent to a 
single given force, is the most common case of resolution. 

11. Representation of Forces Graphically. — The magnittide 
and direction of a force can both be represented by a line ; the 
length of the line representing the magnitude of the force, and 
its direction the direction of the force. 

In order that the length of a line may represent the magni- 
tude of a force, a certain length must be chosen to denote the 
unit force. Then a force of any magnitude will be represented 
by a length which contains the assumed length as many times 
as the magnitude of the given force contains that of the unit 
force. 

In order that the direction of a force may be represented by 
a line, there must be some means of distinguishing between the 
two opposite directions along the line. The usual method is to 
place an arrow-head on the line, pointing in the direction toward 



4 GRAPHIC STATICS. 

which the force acts. If the line is designated by letters placed 
at its extremities, the order in which these are read may indicate 
the direction of the force. Thus, AB and BA represent two 
forces, equal in magnitude but opposite in direction. 

The line of action of a force can also be represented by a line 
drawn on the paper. 

In solving problems in statics, it is usually convenient to 
draw two separate figures, in one of which the forces are 
represented in magnitude and direction only, and in the other 
in line of action only. 

These two species of diagrams will be called force diagra7ns 
and space diagrams, respectively. 

12. Notation. — The use of graphic methods is much facili- 
tated by the adoption of a convenient system of notation in 
the figures drawn. 

There will generally be two figures (the force diagram and 
the space diagram) so related that for every line in one there 
is a corresponding line in the other. 

In the force diagram each line represents a force in inagni- 
t7ide and direction ; in the space diagram the corresponding 
line represents the line of action of the force. These lines 
will usually be designated in the following manner : The line 

denoting the magnitude and direction of 
the force will be marked by two capital 
letters, one at each extremity ; while the 
action-line will be marked by the corre- 
sponding small letters, one being placed at 
-R,. , each side of the line desio:nated. Thus, in 

-t* ig. 1 o 

Fig. I, AB represents a force in magni- 
tude and direction, while its action-line is marked by the letters 
ab, placed as shown. 




COMPOSITION OF CONCURRENT FORCES. 




ITig. S 




The 



§ 2. Composition of CortctLrrent Forces. 

13. Resultant of Two Concurrent Forces. — If two concur- 
rent forces are represented in magnitude and direction by two 
lines AB and BC, their 
resultant is represented 
in magnitude and direc- 
tion by AC. (Fig. 2.) 
Proofs of this proposi- 
tion are given in all 
elementary treatises on 

mechanics, and the demonstration will be here omitted. 
point of application of the resultant is the same as that of the 
given force. Thus if O (Fig. 2) is the given point of applica- 
tion, then ab, be, and ac, drawn parallel to AB, BC, and AC 
respectively, are the lines of action of the two given forces and 
their resultant. The figure marked {A) is a force diagram, and 
{B) is the corresponding space diagram (Art. 11), 

14. Resultant of Any Number of Concurrent Forces. — If any 
number of concurrent forces are represented in magnitude and 
direction by lines AB, BC, CD, . . ., their resultant is repre- 
sented in magnitude and direction by the line AiV, where N is 
the extremity of the line representing the last of the given 
forces. 

This proposition follows immediately from the preceding 
one ; for the resultant of the forces represented by AB and 
BC is a force repre- 
sented by AC; the re- 
sultant oi AC and CD 
is AD, and so on. By 
continuing the process 
we shall arrive at the 
result stated. It is 

readily seen that the order in which the forces are taken does 
not affect the magnitude or direction of the resultant as thus 





FU 




6 GRAPHIC STATICS. 

determined. The point of application of the resultant is the 
same as that of the given forces. Figure 3 shows the force 
diagram and space diagram for a system of four forces repre- 
sented by AB, BCy CD, DE, and their resultant represented 
by AE, applied at the point O. 

It is evident that every system of concurrent forces has for 
its resultant some single force (Art. 9) ; though in particular 
cases its magnitude may be zero. 

15. Force Polygon. — The figure formed by drawing in suc- 
cession lines representing in magnitude and direction any 

number of forces is called a force polygo?i for 
those forces. Thus Fig. 4 is a force polygon 
for any four forces represented in magnitude 
and direction by the lines AB, BC, CD, and 
DEy whatever their lines of action: It may 
happen that the point E coincides with A, in 
which case the polygon is said to be closed. It is evident that 
the order in which the forces are taken does not affect the 
relative positions of the initial and final points. 

I § 3. Equilibrium of Concurrent Forces. 

16. Definition. — A system of forces acting on a body is 
in eqnilibrium if the motion of the body is unchanged by its 
action. 

17. Condition of Equilibrium. — In order that no motion 
may result from the action of any system of concurrent forces, 
the magnitude of the resultant must be zero ; and conversely, 
if the magnitude of the resultant is zero, no motion can 
result. But (Arts. 14 and 15) the condition that the result- 
ant is zero is identical with the condition that the force 
polygon closes. Hence, the following proposition : 

If any system of concurrent forces is in equilibrium, the force 
polygon for the system must close. And conversely, If the force 



EQUILIBRIUM OF CONCURRENT FORCES, 





ir-is 



polygon is closed for any system of concurrent forces^ tJie system 
is in equilibrium,. 

The comparison of this with the analytical conditions of 
equilibrium is given in Art. 22. 

18. Method of Solving Problems in Equilibrium. — If a sys- 
tem of concurrent forces in equilibrium be partially unknown, 
we may in certain cases determine the unknown elements by 
applying the principles of Art. 17. 

The most usual case is that in which two forces are unknown 
except as to lines of action. Thus, suppose a system of five 
forces in equilibrium, 
three being fully 
known, represented in 
magnitude and direc- 
tion by AB, BC, CD 
(Fig. 5), and in lines 
of action by ab, be, cd, 

while concerning the other two we know only their lines of 
action de, ea. To determine these two in magnitude and 
direction, it is necessary only to complete the force polygon of 
which ABCD is the known part. The remaining sides must be 
parallel respectively to de and ea. From D draw a line parallel 
to de, and from A a line parallel to ea, prolonging them till they 
intersect at E. Then DE and EA represent the required forces 
in magnitude and direction, and the complete force polygon is 
ABCDEA. It is evident that ABODE' A is an equally legiti- 
mate form of the force polygon, and gives the same result for 
the magnitude and direction of each of the unknown forces. 
This problem occurs constantly in the construction of stress 
diagrams by the method described in Part II. 

The student will find little difficulty in treating other prob- 
lems in the equilibrium of concurrent forces. 

19. Problems in Equilibrium. — (i) A particle is in equilib- 
rium under the action of five forces, three of which are 



8 GRAPHIC STATICS. 

completely known, while of the remaining two, one is known 
in direction only, and the other in magnitude only. To deter- 
mine the unknown forces. 

(2) Suppose two forces known in magnitude but not in 
direction, the remaining forces being wholly known. 

(3) Suppose one force wholly unknown, the others being 
known. 

§ 4. Resolution of Concurrent Forces. 

20. To Resolve a Given Force into Any Number of Com- 
ponents having the same point of application, we have only to 
draw a closed polygon of which one side shall represent the 
magnitude and direction of the given force ; then the remaining 
sides will represent, in magnitude and direction, the required 
components. This problem is, in general, indeterminate, unless 
the components are required to satisfy certain specified con^ 
ditions. 

[Note. — A problem is said to be indeterminate if its conditions can be satisfied 
in an infinite number of ways. It is determinate if it admits of only one solution. 
Thus, the problem, to determine the values oi x and^ which shall satisfy the equation 
X -\- y — 10, is indeterminate; while the problem, given 2 jr + 3 = 7, to find the 
value of jr, is determinate. The case in which a finite number of solutions is possible 
may be called incompletely determinate. Thus, the problem, given x"^ Ar x — d = o, 
to find X, admits of two solutions, and therefore is incompletely determinate. All 
these classes of problems may be met with in statics.] 

21. To Resolve a Given Force into Two Components. — This 
problem is indeterminate unless additional data are given. For 
if the given force be represented in magnitude and direction by 
a line, any two lines which with the given line form a triangle 
may represent forces which are together equivalent to the given 
force. But an infinite number of such triangles may be drawn. 
The solution of the following four cases of this problem will 
form exercises for the student. In each case the force diagram 
and space diagram should be completely drawn, and the student 
should notice whether the problem is determinate, partially 
determinate, or indeterminate. 



RESOLUTION OF CONCURRENT FORCES. 



(i) Let the lines of action of the required components be 



given. 



(2) Let the two components be given in magnitude only. 

(3) Let the line of action of one component and the magni- 
tude of the other be given. 

(4) Let the magnitude and direction of one component be 



given. 



It will be noticed that these four cases correspond to four 
cases of the solution of a place triangle. 

22. Resolved Part of a Force. — If a force is conceived as 
replaced by two components at right angles to each other, each 
is called the resolved part, "^ in its direction, of the given force. 

It is readily seen that the resolved part of a force repre- 
sented by ^^ (Fig. 6) in the direction of any line XX, is 
represented in magnitude and direction 

A D 

by A^ B\ the orthographic projection of 
AB upon XX. It follows that the 
resolved part (in any given direction) of 
the resultant of any concurrent forces 
is equal to the algebraic sum of the 
resolved parts of its components in that direction ; signs plus 
and minus being given to the resolved parts to distinguish the 
two opposite directions which they 
may have. Thus (Fig. 7) the re- 
solved parts of the forces AB, BC, 
CD, in a direction parallel to XX, 
are A' B' , B' C , CD' ; and their 
algebraic sum is A'D', which is the 
resolved part of the resultant AD. 
If the resultant is zero, D' coincides with A' 

(i) For the equilibrium of any concurrent forces, the sum 
of their resolved parts in any direction must be zero. 




Fi 




Fig. r 



hence, 



* The term "resolute" has been proposed by J. B. Lock (" Elementary Statics ") to 
denote what is here defined as the resolved part of a force. 



lO GRAPHIC STATICS. 

Again, if D^ coincides with A\ then either D coincides 
with A, or else AD is perpendicular to XX \ hence, 

(2) If the sum of the resolved parts of any concurrent 
forces in a given direction is zero, their resultant (if any) 
is perpendicular to that direction. And if the sum of the 
resolved parts is zero for each of two directions, the resultant 
is zero, and the system is in equilibrium. 

Propositions (i) and (2) state the conditions of equilibrium 
for concurrent forces usually deduced in treatises employing 
algebraic methods. 



1 



CHAPTER IL — NON-CONCURRENT FORCES. 

§ I. Composition of Non-concurrent Forces Acting on the Same 

Rigid Body. 

23. Definition of Rigid Body. — A rigid body is one whose 
particles do not change their positions relative to each other 
under any applied forces. No known body is perfectly rigid, 
but for the purposes of statics, most solid bodies may be con- 
sidered as such ; and any body which has assumed a form of 
equilibrium under applied forces, may, for the purposes of 
statics, be treated as a rigid body without error. 

24. Change of Point of Application. — The effect of a force 
upon a rigid body will be the same, at whatever point in its 
line of action it is applied, if only the particle upon which it 
acts is rigidly connected with the body. 

This proposition is fundamental to the development of the 
principles of statics, and is amply justified by experience.* 
In applying the principle, we are at liberty to assume a 
point of application outside the actual body, the latter being 
ideally extended to any desired limits. 

25. Resultant of Two Non-Parallel Forces. — If two com- 
planar forces are not parallel, their lines of action must inter- 
sect, and the point of intersection may be taken as the point 
of application of each force. Hence, they may be treated as 



*This proposition may be proved analytically by deducing the equations of motion 
of a rigid body, and showing that the effect of any force on the motion of the body 
depends only upon its magnitude, direction, and line of action. But such a proof is, of 
course, outside the scope of this work. 

TI 



12 



GRAPHIC STATICS. 



concurrent forces, and their resultant may be determined as 
in Art. 13. The following proposition may therefore be 
stated : 

If two forces acting in the same plane on a rigid body 
are represented in magnitude and direction by AB and BC, 
their resultant is represented in magnitude and direction by 
AC, and its line of action passes through the point of inter- 
section of the lines of action of the giv^en forces. Its point 
of application may be any point of this line. 

It may happen that the point of intersection of the two 
given lines of action falls outside the limits available for the 
drawing. In such a case it will be most convenient to find 
the resultant by the method to be explained in Art. 27. The 
same remark applies to the case of two parallel forces. 



26. Resultant of Any Number of Non-concurrent Forces — 
First Method. — ^The method of the preceding article may be 
extended to the determination of the resultant of any number 
of forces acting on the same rigid body. Let AB, BC, CD, DE 

(Fig. 8), represent in 
magnitude and direction 
four forces, and let ab, 
be, cd, de represent their 
lines of action. To find 
their resultant, we may 
proceed as follows : 
The resultant of AB 
and BC is represented 
in magnitude and direction by AC, and in line of action by ac 
drawn parallel to AC through the point of intersection of ab 
and be. Combining this resultant with CD, we get as t/ieh' 
resultant a force represented in magnitude and direction by 
AD, and in line of action by ad drawn parallel to AD through 
the point of intersection of ae and ed. This is evidently the 
resultant of AB, BC, and CD. In the same way, this resultant 




COxMPOSITION OF XON-COx\CURRENT FORCES. 13 

combined with DE o-ives for their resultant a force whose mas-- 
nitude and direction are represented by AE, and whose line of 
action is ac^ parallel to AE and passing through the point in 
which ad intersects de. This last force is the resultant of the 
four given forces. 

The process may evidently be extended to the case of any 
number of forces. 

As in the case discussed in the preceding article, this method 
will become inapplicable or inconvenient in case any of the 
points of intersection fall outside the limits available for the 
^drawing. For this reason it is usually most convenient to 
employ the method described in Art. 27. 

The student should bear in mind that the length and direc- 
tion AE and the line ae are not the magnitude, direction, and 
line of action of any actual force applied to the body. By the 
resultant is meant an ideal force, which, if it acted, would 
produce the same effect upon the motion of the body as is 
produced by the given forces. It is a force which may be 
conceived to replace the actual forces, and may be assumed to 
be applied to any particle in its line of action, provided that 
particle is conceived as rigidly connected with the given body. 
The line of action may in reality fail to meet the given body. 
(See Art. 24.) 

2j. Resultant of Non-concurrent Forces — Second Method. — 

This method will be described by reference to an example. 
Referring to Fig. 9, let AB, BC, CD, DE represent in magni- 
tude and direction four forces whose lines of action are ab, be, 
ed, de ; and let it be required to find their resultant. Draw the 
force polygon ABCDE, and from any point O in the force 
diagram draw lines OA, OB, OC, OD, OE. These lines may 
represent, in magnitude and direction, components into which 
the given forces may be resolved. Thus AB is equivalent to 
forces represented by ^6^ and OB acting in any lines parallel 
to AO, OB, whose point of intersection falls upon ab ; BC is 



14 



GRAPHIC STATICS. 



equivalent to forces represented by BO, OC, acting in any lines 
parallel to BO, OC, which intersect on be ; and so for each of 
the given forces. The four given forces may, therefore, be 
replaced by eight forces given in magnitude and direction by 




AO, OB, BO, OC, CO, OD, DO, OE, with proper lines of 
action. Now, it is possible to make the lines of action of the 
forces represented by OB and BO coincide ; and the same is 
true of each pair of equal and opposite forces, OC, CO ; OD, 
DO. To accomplish this, let AO, OB act in lines ao, ob, inter- 
secting at any assumed point of ab. Prolong ob to intersect 
be, and take the point thus determined as the point of inter- 
section of the lines of action of BO, OC ; these lines are then 
bo, oe. Similarly prolong oe to intersect ed, and let the point 
of intersection be taken as the point at which CD is resolved 
into CO and OD ; the lines of action of these forces are then 
eo and od. In like manner choose do, oe, intersecting on de, as 
^the lines of action of DO, OE. If this is done, the forces OB, 
BO will neutralize each other and may be omitted from the 
system ; also the pairs OC, CO, and OD, DO. Hence, there 
remain only the two forces represented in magnitude and 
direction by AO, OE, and in lines of action by ao, oe. 
Their resultant is given in magnitude and direction by AE, 
and its line of action is ae, drawn parallel to AE through the 



COMPOSITION OF NON-CONCURRENT FORCES. 



15 



point of intersection of oa and oe ; and this is also the result- 
ant of the given system. 

By carefully following through this construction the student 
will be able to reduce it to a mechanical method, which can 
be readily applied to any system. 

28. Funicular Polygon. — The polygon whose sides are oa, ob, 
oc, ody oe, is called 2, fiinicidar polygon^ for the given forces. 

Since the point at which the two components of AB are 
assumed to act may be taken anywhere on the line ab, there 
may be any number of funicular polygons with sides parallel 
to oa, ob, etc. Again, if O is taken at a different point, there 
may be drawn a new funicular polygon starting at any point 
of ab ; and by changing the starting point any number of 
funicular polygons may be drawn with sides parallel to the 
new directions of OA, OB, etc. Moreover, different force 
and funicular polygons may be obtained by changing the order 
in which the forces are taken. 

It may be proved geometrically that for every possible funic- 
ular polygon drawn for the same system of forces, the last 
vertex, determined by the above method, will lie on the same 
line parallel to the closing side of the force polygon (as ae, 
Fig. 9). Such a proof is outside the scope of this work. 
The truth of the proposition may, however, be shown from 
the principles of statics. For if it were not true, it would 
be possible by the above method to find two or more forces, 
having different lines of action, which are equivalent to each 
other, because each is equivalent to the given system. But 
this is impossible. 

29. Examples. — i. Choose five forces, assigning the magni- 
tude, direction, and line of action of each, and find their 
resultant by constructing the force and funicular polygons. 

2. Draw a second funicular polygon, using the same point 
O in the force diagram. 

*Also called equilibrium polygon. 



l6 GRAPHIC STATICS. 

3. Draw a third funicular polygon, choosing a new point O. 

4. Solve the same problem, taking the forces in a different 
order. 

30. Definitions. — The point O (Fig. 9) is called the /<?/<:' of 
the force polygon. The lines drawn from the pole to the 
vertices of the force polygon may be called rays. The sides 
of the funicular polygon are sometimes called strings. 

Each ray in the force diagram is parallel to a corresponding 
string in the space diagram. As a mechanical rule, it should 
be remembered that the two rays drawn to the extremities of the 
line representing any force are respectively parallel to the tzuo 
strings which intersect on the line of actioji of that force. 

The rays terminating at the extremities of any side of the 
force polygon represent in magnitude and direction two com- 
ponents that may replace the force represented by that side ; 
while the corresponding strings represent the lines of action of 
these components. Thus (Fig. 9) the force represented by BC 
may be replaced by two forces represented by BOy OCj acting 
in the lines bo, oc. 

The pole distance of any force is the perpendicular distance 
from the pole to the line representing that force in the force 
diagram. It may evidently be considered as representing the 
resolved part, perpendicular to the given force, of either of the 
components represented by the corresponding rays. Thus, OM 
(Fig. 9) is the pole distance of AB ; and OM represents the 
resolved part, perpendicular to AB, of either OA or OB. The 
student should notice particularly that the pole distance repre- 
sents di force inagnitnde and not a length. 

31. Forces not Possessing a Single Resultant. — It may hap- 
pen that the first and last sides of the funicular polygon are 
parallel, so that the above construction fails to give the line of 
action of the resultant. This will be the case if the pole is 
chosen on the line AE (Fig. 9), because the first and last sides 
of the funicular polygon are respectively parallel to OA and 



COMPOSITION OF NON-CONCURRENT FORCES. \y 

OE. The difficulty will, in this case, be avoided by taking the 
pole at some point not on the line AE. But in one particular 
case ' OA and OE will be parallel, wherever the pole be taken. 
By inspection of the force diagram it is seen that this can 
occur only when the point E coincides with A. In this case 
AO and OE represent equal and opposite forces ; and unless 
their lines of action coincide, they cannot be combined into a 
simpler system. If their lines of action are coincident, the 
forces neutralize each other and the resultant is zero ; if not, 
the system reduces to a couple (Art. 7). If the lines of action 
oi AO and OE coincide, they may still be regarded as forming 
a couple, its arm being zero. Therefore, 

1/ tJie force polygon for any system of forces closes, the result- 
ant is a couple. 

Now, it is evident that by shifting the starting point for the 
funicular polygon, the lines of action oi AO and OE will be 
shifted ; and by taking a new pole, their magnitude, or direc- 
tion, or both, may be changed. Hence, there may be found 
any number of couples, each equivalent to the same system 
of forces, and therefore equivalent to each other. The relation 
between equivalent couples is discussed in Art. 52. 

Example. — Assume five forces whose force polygon closes, 
the lines of action being taken at random. Draw two funicular 
polygons, using the same pole, and a third, using a different 
pole ; thus reducing the given system to an equivalent couple 
in three different ways. 

32. Resultant Force and Resultant Couple. — From the above 

discussion (Arts. 27 and 31), it is evident that any system of 
complanar forces, acting on the same rigid body, is equivalent 
either to a single force or to a couple. In other words, every 
system of complanar forces possesses either a resultant force or a 
resultant couple. (See Art. 9.) 

33. Comparison of Methods. — Of the methods given in Arts. 
26 and 27, for finding the resultant of a system of non-concur- 



1 8 GRAPHIC STATICS. 

rent forces, the first is a special case of the second. For if the 
pole in Fig. 9 be chosen at the point A, and the first string be 
made to pass through the point of intersection of ad and dc, the 
construction becomes identical with that of Fig. 8. If the first 
method is employed, we are liable to meet the difficulty men- 
tioned in Arts. 25 and 26, that some of the required points of 
intersection do not fall within convenient limits. 

In the second method, since the pole may be chosen at pleas- 
ure, the rays may generally be caused to make convenient angles 
with the given forces. It will rarely be necessary that the rays 
intersect the corresponding force lines at acute angles, and will 
never be necessary that a ray shall be parallel to the force of 
which it represents a component. Hence the pole may gener- 
ally be so chosen that all intersections shall fall within conven- 
ient limits. The difficulty above mentioned may therefore 
usually be avoided by using the method last described. This 
method is especially convenient in the case of parallel forces. 

Example. — Choose six parallel forces and find their resultant 
by constructing the force and funicular polygons. 

34. Closing of the Funicular Polygon. — Let the force poly- 
gon be drawn for any given forces, and let A and E be the 
initial and final points. Suppose the funicular polygon drawn, 
corresponding to any pole O. The given system is equivalent 
to two forces represented in magnitude and direction by AO, 
OE, their lines of action being the first and last sides of the 
funicular polygon. In general, these lines of action will not be 
parallel; but, as explained in Art. 31, it may happen that they 
are parallel. And if parallel, it may happen that they coincide. 
In this case, the funicular polygon is said to be closed. 

§ 2. Equilibrium of Non-concurrent Forces. 

35. Conditions of Equilibrium. — From the preceding articles 
it follows that, in order that a system of complanar forces may 
be in equilibrium, the following conditions must be satisfied : 



EQUILIBRIUM OF NON-CONCURRENT FORCES. iq 

{a) The force polygon must close ; otherwise the construction of 
Art. 27 will always lead to a resultant force of finite magnitude. 

{b) Every funicular polygon Duist close; otherwise, by the 
method of Art. 27, the system can be reduced to two forces with 
different lines of action ; the resultant cannot, therefore, be zero. 

Conversely, if tJie force polygon is closed and one funicular 
polygon also closes, tJie system is in equilibrium. For, the clos- 
ing of the force polygon shows that the system may be reduced 
to two equal and opposite forces (Art. 31) ; and the closing of 
the funicular polygon shows that these have the same line of 
action, and therefore balance each other. 

It follows that if the force polygon and one funicular polygon 
are closed, all funicular polygons will clo'se. 

36. Auxiliary Conditions of Equilibrium. — If any system of 
forces in equilibrium be divided into two groups, the resultants 
of the two groups are equal and opposite and have the same 
line of action. 

Particular case. — If three forces are in equilibrium, their 
lines of action must meet in a point, or be parallel. For, if 
the lines of action of two of the forces intersect, their result- 
ant must act in a line passing through the point of inter- 
section. But this resultant must be equal and opposite to 
the third force and have the same line of action. 

37. General Method of Solving Problems in Equilibrium. — 

The principles of Art. 35 furnish a general method of solv- 
ing, graphically, problems relating to the equilibrium of com- 
planar forces acting on a rigid body. The problems to be 
solved will always be of the following kind : 

A body is in equilibrium under the action of forces, some 
of which are completely known, and others wholly or partly 
unknown. It is required to determine the unknown elements. 
The required elements may be either the magnitudes or the 
lines of action of the forces. 



20 



GRAPHIC STATICS. 



The general method of procedure is as follows : First, draw 
the force polygon and funicular polygon so far as possible 
from the given data ; then complete them, subject to the 
condition that both polygons must close. 

This general method will be illustrated by the solution of 
several problems of frequent occurrence. We may here meet 
with both determinate and indeterminate problems (Art. 20). 
In order that a problem may be determinate, the given data, 
together with the condition that the force polygon and funic- 
ular polygon are to close, must be just sufficient to fully 
determine these figures. There are many possible cases fur- 
nishing determinate problems. Some of these cannot readily 
be solved graphically. In the following articles are treated 
three important cases to which the general method above 
outlined is well adapted. 



a 



d d 



38. Problems in Equilibrium. I. — A rigid body is in equi- 
librium under the action of a system of parallel forces, all 
known except two, these being unknown in magnitude and 
direction, but having known lines of action. It is required 
to fully determine the unknown forces. 

Let the known lines of action of five forces in equilibrium 

be ab, bc^ cd, de, ea (Fig. 10), and let 
AB, BC, CD represent the known 
forces in magnitude and direction, 
while DE and EA are at first un- 
known. The force polygon, so far 
as it can be drawn from the given 
data, is the straight line ABCD. 
Choose a pole O and draw rays OA, 
OB, OC, OD. Parallel to these draw 
in the space diagram the strings oa, 
obj OC, od, four successive sides of the 
funicular polygon. This much can 
be drawn from the data given. We must now complete the 



TTig. 10 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 21 

two polygons and cause both to close. In the funicular poly- 
gon, but one side remains to be drawn ; and in the force 
polygon but one vertex remains unknown. If the forces are 
taken in the order AB, BC, CD, DE, EA, the unknown vertex 
in the force polygon is the one to be marked E ; the unknown 
side of the funicular polygon is oe, and is to be parallel to 
the line OE. But the string oe must pass through the inter- 
section of od with de, and also through the intersection of 
oa with ac. Hence, since these intersections are both known, 
oc can be drawn at once as shown in the figure ; and then 
OE can be drawn parallel to oe. This fixes the point E ; 
and DE and EA represent, in magnitude and direction, the 
required forces whose lines of action are de, ea. 

39. Problems in Equilibrium. II. — A rigid body is in 
equilibrium under the action of a system of non-parallel forces, 
all known, except two ; of these, the line of action of one 
and the point of application of the other are given. It is 
required to completely determine the unknown forces. 

Let the system consist of five forces to be represented in 
the usual way ; in magnitude and direction by lines marked 
AB, BC, CD, DE, EA ; and in lines of action by ab, be, cd, 
de, ea. Of these lines let be, ed, de, ea (Fig. 11), be given, and 
let Z be a given point of ab. Also let BC, CD, and DE be 
known, while EA, AB are unknown. First, draw the force 
polygon as far as possible, giving B, C, D, E, four consecutive 
vertices of the force 

polyo'on. The side EA D , ^r 

must be parallel to ea, r ^^^-^X ^^\ ii---^ n 

but its length is un- i ^^m )q \ |\ ^^>f^^ 

known, hence the ver- bv" i / i/ \^/^ 

tex A or the force \^ | / / 'c X 

polygon cannot be ^^ y^c ^V 

fixed. Next, draw the ^. „ 

Fig. 11 

funicular polygon so 



22 GRAPHIC STATICS. 

far as possible from the given data. Choose the pole O, and 
draw the rays OB, OC, OD, OB. The remaining ray, OA, 
cannot yet be drawn. Now, in the funicular polygon, the 
sides od, oc must intersect on be ; oc and od must intersect 
on cd\ <?<3f and ^^ must intersect on de \ oe and oa must inter- 
'sect on ea ; and oa and ob must intersect on ab. Since L 
is the only known point of the line ab, let this point be 
taken as the point of intersection of oa and ob. Now draw 
ob through L parallel to OB, and successively oc, od, oe, par- 
allel to OC, OD, OE. We may now draw oa, joining L with 
the point in which oe intersects ea. This completes the funic- 
ular polygon. The force polygon can now be completed as 
follows : From O draw a line parallel to oa and from E a line 
in the known direction of EA ; their intersection determines 
A. This determines both EA and AB, and the force polygon 
is completely known. The line of action ab may now be 
drawn through L parallel to AB, and the forces are completely 
determined. 

40. Problems in Equilibrium. III. — A rigid body is in 
equilibrium under the action of any number of forces, of 
which three are known only in line of action ; the remaining 
forces being completely known. It is required to determine 
the unknown forces. 

Let the given forces be six in number, their lines of action 
being represented by ab, be, ed, de, ef, fa (Fig. 12), and let AB, 
BC, and CD be known, while the remaining three forces are un- 
known in magnitude and direction. The known data make it 
possible to draw at once three sides of the force polygon, 
namely, AB, BC, CD; and the four rays OA, OB, OC, OD, 
from any assumed pole O. Also, four sides of the funicular 
polygon, oa, ob, oe, od, may be drawn at once. But oe and of 
are unknown ; also DE, EF, and FA in the force polygon. 

Since any side of the funicular polygon can be drawn through 
any chosen point, let the polygon be started by drawing oa 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



^?> 



through the intersection of ^and fa. We may then draw suc- 
cessively ob, oc, od. Now, of is unknown in direction ; but it 
is to be drawn through the intersection of oa and fa, hence, 
whatever its direction, it intersects ef in the same point (since 
oa was drawn through the intersection of ef and fa). Hence, 
the two vertices of the funicular polygon falling on ef and fa 
coincide at the intersection of these two lines. We may there- 




fore draw oe through this point and also through the point 
already found by the intersection of od and de. Now draw a 
line from O parallel to oe, and from D a line parallel to de ; 
their intersection is E. Draw from E a line parallel to ef and 
from A a line parallel to fa ; their intersection determines F. 
The force polygon is now completely drawn, and DE, EF, FA 
represent, in magnitude and direction, the required forces. The 
remaining ray OF and the corresponding string of may now be 
drawn, but are not needed. 

The construction might have been made equally well by 
choosing the intersection of de and ef as the starting point, 
since two vertices of the funicular polygon may be made to 
coincide at that point. Or, the point of intersection of de and 
af might be chosen ; but in that case, the order of the forces 
should be so changed as to make de and af consecutive. If 
this were done the figures should be relettered to agree with 



24 



GRAPHIC STATICS. 



the order in which the forces were taken. It may be noticed 
that the directio7i of the string first drawn may be chosen arbi- 
trarily and the pole so taken as to correspond to the direction 
chosen. This is important in the treatment of the following 
special case. 

Case of inaccessible points of intersection. — It may happen 
that the lines of action de, ef and fa have no point of intersec- 
tion within convenient limits. When this is the case, the 
method just given may still be applied, but involves the geo- 
metrical problem of drawing a line through an inaccessible point. 
For example, if ef and fa intersect beyond the limits of the 
drawing, as shown in Fig. 13, we may proceed as follows: 
Choose some point of ab and from it draw a line through the 
point of intersection of ef and fa. (This can be done by a 
method to be explained presently.) Let this line be oa. From 
the known point A of the force polygon draw a line parallel to 





Fig. 13 



oa and choose a pole upon it. Draw the rays OB, OC, OD ; 
then the corresponding strings in the order ob, oc, od. From 
the point in which od intersects de draw a line to the inaccessible 
point of intersection of ef and fa ; this will be oe. The force 
polygon can now be completed just as in the preceding case. 

A line may be drawn through the inaccessible point of inter- 
section of two given lines by the following method : Let AC, 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 



25 



BD (Fig. 14), be the given lines, and let it be required to draw 

a line through their point of intersection from some point P. 

Draw PA intersecting AC and 

BD, and from C, any point of 

AC, draw a line CQ parallel to 

PA. From E, a point of BD, 

draw EA, EP ; also draw CF 

parallel to AE, and FQ parallel 

to EP. Then PQ is the line 

required. For, by the similar 

triangles, it is easily shown that 

AB CG 

BP 



-^ 



F D 






( — — o 



Q 

Fig. 14= 



; which proves that AC, BG, and PQ meet in a point. 



GQ 

Exception. — If the lines of action of the three unknown 
forces meet in a point (or are parallel), the problem is impos- 
sible of solution, unless it happens that the resultant of the 
known forces acts in a line through the point of intersection 
of those three lines of action (or parallel to them) ; in which 
case the problem is indeterminate. These cases will not be 
discussed here. 

Remark. — In the problems treated in this and the two pre- 
ceding articles, it will be noticed that the forces should be taken 
in such an order that those which are completely known are 
consecutive. Otherwise the number of unknown lines in the 
force and funicular polygons will be increased. [For an- 
other method of solving this problem, see Levy's " Statique 
Graphique."] 

41. Examples. — i. A rigid beam AB rests horizontally 
upon supports at A and B, and sustains loads as follows : 
Its own weight of 100 lbs. acting at its middle point ; a load 
of 50 lbs. at (^ ; a load of 60 lbs. at D ; and a load of 80 lbs. 
at E. The successive distances between A, C, D, E, and B 
are 4 ft., 6 ft., 7 ft., and 10 ft. Required the upward pressures 
on the beam at the supports. 



26 GRAPHIC STATICS. 

2. A rigid beam AB is hinged at A, and rests horizontally 
with the end B upon a smooth horizontal surface. The beam 
sustains loads as in Example i, and an additional force of 40 
lbs. is applied at the middle point at an angle of 45° with the 
bar in an upward direction. Required the pressures upon the 
beam at A and B. [The pressure at A may have any direc- 
tion, while the pressure at B must be vertically upward, i.e., 
at right angles to the supporting surface. Hence, this is a 
particular case of Problem II.] 

3. Let the end B of the bar rest against a smooth surface 
making an angle of 30° with the horizontal ; the remaining 
data being as in Example 2. 

4. A rigid bar 2 ft. long weighs 10 lbs., its center of gravity 
being 8 inches from one end. The bar rests inside a smooth 
hemispherical bowl of 15 inches radius. What weight must 
be applied at the middle point in order that the bar may rest 
when making an angle of 15° with the horizontal.'* Also, 
what are the reactions at the ends ? 

5. A uniform bar 20 inches long, weighing 10 lbs., rests 
with one end against a smooth vertical wall and the other 
end overhanging a smooth peg 10 inches from the wall. A 
weight P is suspended from the end so that the bar is in 
equilibrium when making an angle of 30° with the horizontal. 
Find P, and the pressures exerted on the bar by the wall 
and peg. 

42. Special Methods. — Certain problems can be treated 
more simply by other methods than by the general method 
of constructing the force and funicular polygons. This is 
sometimes true of the following : 

Problem. — A rigid body is held in equilibrium by four 
forces acting in known lines, only one being known in magni- 
tude and direction. It is required to completely determine 
the three remaining forces. (See Fig. 15.) 

Let the four forces have lines of action ab, be, cd, da, and let 



EQUILIBRIUM OF NON-CONCURRENT FORCES. 27 

AB be drawn representing the magnitude and direction of the 
known force. Now the resultant of the forces whose lines 
of action are da and ab must act in a line passing through 
My the point of intersection of these lines ; and the resultant 
of the other two forces must act in a line passing through 
iV, the point of intersection of be and cd. But the two 



resultants must be equal and opposite and have the same 
line of action, else there could not be equilibrium. (Art. 36.) 
Hence, each must act in the line MN. Draw BD parallel 
to MN, and AD parallel to ad\ the point D being determined 
by their intersection. Then DA represents in magnitude and 
direction the force acting in da, and DB the resultant of 
DA and AB. But BD must represent the resultant of the 
two remaining forces ; hence these two forces are represented 
by BC and CD drawn from B and D parallel respectively to 
be and cd. 

This problem is a special case of that treated in Art. 40. 
But the construction here given will in many cases be the 
simpler one. 

Example. — A rigid body has the form of a square ABCD, 
the side AB being horizontal, and BC vertical. The weight 
of the body is 100 lbs., its center of gravity being at the inter- 
section of the diagonals. It is held in equilibrium by three 
forces as follows : P^ acting at C in the line AC ; Po_ acting at 
D in the line AD ; and P^ applied at B and acting in the line 
joining B with the middle point of AD. Required to com- 
pletely determine P^, P<y, and P^. 



28 GRAPHIC STATICS. 

§ 3. Resolution into Non-concntTent Systems. 

43. To Replace a Force by Two Non-concurrent Forces. — 

This may be done in an infinite number of ways. The lines 
of action of the two components must intersect at a point on 
the line of action of the given force, and they must further 
satisfy the same conditions as concurrent forces. (Art. 21.) 

44. To Replace a Force by More Than Two Non-concurrent 
Forces. — This may be done by first resolving the given force 
into two forces by the preceding article^ and then resolving 
one or both of its components in the same way. This problem 
and that of Art. 43 are indeterminate. (See note, Art. 20.) 
To make such a problem determinate, something must be 
specified regarding the magnitudes and lines of action of the 
required components. We shall consider some of the particular 
cases which are of frequent use in the treatment of practical 
problems. 

45. Resolution of a Force into Two Parallel Components. — 

Let it be required to resolve a force into two components 
having given lines of action parallel to its own. 

If the given force be reversed in direction, it will form with 
the required components a system in equilibrium. The com- 
ponents may then be determined by the method of Art. 38. 

Example. — Let the student solve two particular cases of 
this problem, taking the line of action of the given force (i) 
between those of the components, (2) outside those of the 
components. 

46. Resolution of a Force into Three Components. ~Pr^<5/^w. 

— To resolve a force into three components having known lines 
of action. 

If the given force be reversed in direction, it will form, with 
the required forces, a system in equilibrium. Hence these 
forces may be determined by either of the two methods given 



RESOLUTION INTO NON-CONCURRENT SYSTEMS. 



29 



in Arts. 40 and 42. Or, the following reasoning may be 
employed, leading to the same construction as that of Art. 42. 

Let AB (Fig. 16) represent the given force in magnitude 
and direction, and a^/ its line of action ; the lines of action of 
the components being given as al?, be, cd. Since the given 
force may be assumed to act at any point in the line ad, let its 
point of application be taken at M, the point of intersection of 





Fig. 16 



ad and cd. Resolve it into two components acting in the lines 
cd and MN, N being the point of intersection of ab and be. 
These components are represented in magnitude and direction 
by AC, CD. Let AC, acting in the line J/^V (also marked ac), 
be resolved into components having lines of action ab, be. 
These components are given in magnitude and direction by 
AB, BC, drawn parallel respectively to ab, be. Hence, the 
given force, acting in ad, is equivalent to the three forces 
represented in magnitude and direction by AB, BC, CD, acting 
in the lines ab, be, cd. 

If the line of action of the given force does not intersect any 
one of the given lines ab, be, cd, within the limits of the draw- 
ing, it may be replaced by two components ; then each may be 
resolved in accordance with the above method, and the results 
combined. If the three lines ab, be, cd, are all parallel to ad, 
or if the four lines intersect in a point, the problem is inde- 
terminate. This is evident from the preceding articles, since, 
by methods already given, a part oi AD can be replaced by 
two components acting in any two of the given lines, as ab, 
be ; and the remaining part by two components acting in ab, cd; 



30 



GRAPHIC STATICS. 



or in be, cd. This construction evidently admits of infinite 
variation. 

For another method of solving the above problem, see Clarke's 
*' Graphic Statics," p. i6. 

§ 4. Moments of Forees and of Couples. 

47. Moment of a Force. — Definition. — The moment of a 
force with respect to a point is the product of the magnitude of 
the force into the perpendicular distance of its line of action 
from the given point. The moment of a force with respect to 
an axis perpendicular to the force is the product of the magni- 
tude of the force by the perpendicular distance from the axis 
to the line of action of the force. 

If the moment is taken with respect to a point, that point is 
called the origin of moments. The perpendicular distance from 
the origin, or axis, to the line of action of the force is called 
the arm. 

Rotation tendency of a force. — The moment of a force 
measures its tendency to produce rotation about the origin, or 
axis. Thus, if a rigid body is fixed at a point, but free to turn 
about that point in a given plane, any force acting upon it in 
that plane will tend to cause it to rotate about the fixed point. 
The amount of this tendency will be proportional both to the 
magnitude of the force and to the distance of its line of action 
from the given point ; that is, to the moment of the force with 
respect to the point, as above defined. 

Rotation in any plane may have either of two opposite 
directions, which may be distinguished from each other by signs 
plus and minus. Rotation with the hands of a watch supposed 
placed face upward in the plane of the paper will be called 
negative, and the opposite kind positive. It would be equally 
legitimate to adopt the opposite convention, but the method 
here adopted agrees with the usage of the majority of writers. 

The sign of the moment of a force is regarded as the same 
as that of the rotation it tends to produce about the origin. 






MOMENTS OF FORCES AND OF COUPLES. 31 

Moment represented by the area of a triangle. — If a triangle 
be constructed having for its vertex the origin of moments, and 
for its base a length in the line of action of a force, numerically 
equal to its magnitude, then the moment of the force is numeri- 
cally equal to double the area of the triangle. This follows at 
once from the definition of moment. 

48. Moment of Resultant of Two Non-parallel Forces. — Propo- 
sitio?i. — The moment of the resultant of two non-parallel forces 
with reference to a point in their plane is equal to the algebraic 
sum of their separate moments with reference to the same point. 

In Fig. 17, let AB, BC, and AC represent in magnitude and 
direction two forces and their resultant ; and let ab, be, ac be 
their lines of action, intersecting; in a .,_ -^ , 

point N. Choose any point M as the \ '\\""^ V 

origin of moments. Y^-d^y o^ NP = AB\ \ ^.^^^--r,---}^^^ 

NQ = BC; and NR = AC. Then the V'^''J^\y^P/ 
moments of the three forces are re- k,\ /^"^^'W / 

spectively equal to double the areas of ^^yjv^,-—"" P bQ 
the triangles MNP, MNQ, MNR. ' ^ ' ^^^ 

These three triangles have a common ^^ / 

side MN, which may be considered the A e 

base ; hence their areas are propor- ■^^^* ^"^ 

tional to their altitudes measured from that side. These alti- 
tudes are PP\ QQ\ PR', perpendicular to MJV. Now, if PP'^ 
is parallel to MN, P"R' is equal to PP' ; also RP" equals 
QQ' (since they are homologous sides of the equal triangles 
NQQ', PRP"). Hence RR' = PP' + QQ\ and therefore 

Area J/iV7? = area JfTVP + area MNQ ; 

which proves the proposition. 

If the origin of moments is so taken that the moments of 
AB and BC have opposite signs, the demonstration needs modi- 
fication. The student should attempt the proof of this case for 
himself. 



32 GRAPHIC STATICS. 

49. Moment of a Couple. — Definition. — The moment of a 
couple about any point in its plane taken as an origin is the 
algebraic sum of the moments of the two forces composing it 
with reference to the same origin. 

Proposition, — The moment of a couple is the same for every 
origin in its plane, and is numerically equal to the product of 
the magnitude of either force into the arm of the couple. 
(Art. 7.) 

The proof of this proposition can readily be supplied by the 
student. 

50. Moment of the Resultant of Any System. — Proposition, 
— The algebraic sum of the moments of any given complanar 
forces, with reference to any origin in their plane, is equal to 
the moment of their resultant force or resultant couple with 
reference to the same origin. 

The construction employed in proving this proposition is 
similar to that used in Art. 27, which the student may profit- 
ably review at this point. Referring to Fig. 9, let the given 
forces be represented in magnitude and direction by AB, BCy 
CD^ DE, and in lines of action by ab, be, eel, de. Let the 
origin of moments be any point in the space diagram. As in 
Art. 27, replace AB by AO, OB, acting in lines ao, ob\ replace 
BC by BOy OC, acting in lines bo, oe ; replace CD by CO, OD, 
acting in lines eo, od; and replace DE by DO, OE, acting in 
lines do, oe. (For brevity, we refer to a force as AB, meaning 
"the force represented in magnitude and direction by AB.'') 

Now by Art. 48, we have, whatever the origin. 

Moment of y4i5 = moment of ^(9 + moment of OB', 
" BC^ " " BO^ '' " OC', 
<' '' CD=^ " '' CO^ " " OD; 

" DE= " ** D0-{- '' '' OE. 

Since the forces represented by BO and OB have the same 
line of action, their moments are numerically equal but have 



MOMENTS OF FORCES AND OF COUPLES. 33 

opposite signs ; and similar statements are true of CO and 
OC, and of DO and OD. Hence, the addition of the above 
four equations shows that the sum of the moments of AB, 
BC, CDy DE, is equal to the sum of the moments of AO 
and OE. Now the given system has either a resultant force 
or a resultant couple. In the first case the resultant of the 
system is the resultant of AO and OE, and its moment is 
equal to the algebraic sum of their moments, by Art. 48. 
In the second case (which occurs only when E coincides with 
A), the resultant couple is composed oi AO and OE (which 
in this case are equal and opposite forces), and the moment 
of the couple is, by definition, equal to the algebraic sum of 
the moments of AO and OE. Hence, in either case, the 
proposition is true. 

It should be noticed that the proof here given applies to 
systems of parallel forces, as well as to non-parallel systems. 
The proposition of Art. 48 could be extended to the case of 
any number of forces, by considering first the resultant of 
two forces, then combining this resultant with the third force, 
and so on ; but the method would fail if, in the process, it 
became necessary to combine parallel forces. The method 
here adopted is not subject to this failure. 

51. Condition of Equilibrium. — It follows from what has 
preceded, that if a given system is in eguilibritnn, the alge- 
braic sum of the moments must be zero, whatever the origin. 
For, in case of equilibrium, AO and OE (Fig. 9) are equal and 
opposite and have the same line of action ; hence, the sum 
of their moments (which is the same as the sum of the 
moments of the given forces) is equal to zero. Conversely, 
If the algebraic sum of the fttoments is zero for eveiy origin, 
the system must be in equilibrium. For, if it is not, there 
is either a resultant force or a resultant couple. But the 
moment of a force is not zero for any origin not on its line 
of action ; and the moment of a couple is not zero for any 



34 



GRAPHIC STATICS. 



origin. For a fuller discussion of the conditions of equilibrium, 
see Arts. 58 and 59. 

52. Equivalent Couples. — Proposition. — If a system has for 
its resultant a couple, it is equivalent to any couple whose 
moment is equal to the sum of the moments of the forces 
of the system. 

For, as already seen (Art. 31), when the resultant is a 
couple, the force polygon is closed. Let the initial and final 
points of the force polygon coincide at some point A^ and 
let O be the pole. Then the forces of the resultant couple 
are represented in magnitude and direction by AO^ OA. Since 
the position of O is arbitrary, the force AO (or OA) may be 
made anything whatever in magnitude and direction. Also 
the line of action of the force AO may be taken so as to pass 
through any chosen point. Hence, the resultant couple may 
have for one of its forces any force whatever in the plane of 
the given system ; and the other force will have such a line 
of action that the moment of the couple will be equal to the 
sum of the moments of the given forces. 

This reasoning is equally true if the given system is a 
couple. Hence, a couple is equivalent to any other couple 
having the same moment. In other words, all couples whose 
^nonients are equal are equivalent ; and conversely, all equiva- 
lent couples have equal moments. 

[Note. — The construction above discussed fails if all forces of the given system 
are parallel to the direction chosen for the forces of the resultant couple. For then 
the force polygon is a straight line, and if the pole is chosen in that line, the 
strings of the funicular polygon are parallel to the lines of action of the forces, and 
the polygon cannot be drawn. But in this case, the system may first be reduced to a 
couple whose forces have some other direction, and this couple may be reduced to 
one whose forces have the direction first chosen. Hence, the proposition stated 
holds in all cases.] 

53. Moment of a System. — Definition. — The moment of a 
system of forces is the algebraic sum of the moments of the 
forces of the system. 



GRAPHIC DETERMINATION OF MOMENTS. 35 

54. Moments of Equivalent Systems. — Proposition. — The 
moments of any two equivalent systems of complanar forces 
with respect to the same origin are equal. 

This follows immediately from the preceding articles. For, 
if the two systems are equivalent, each is equivalent to the 
same resultant force or resultant couple, and the moments of 
the two systems are therefore each equal to the moment of 
this resultant and hence to each other. 

§ 5. Graphic Determination of Moments. 

55. Proposition. — If, through any point in the space dia- 
gram a line be drawn parallel to a given force, the distance 
intercepted upon it by the two strings corresponding to that 
force, multiplied by the pole distance of the force, is equal 
to the moment of the force with respect to the given point. 

By the strings ''corresponding to" a given force are meant 
the two strings which intersect at a point on its line of action. 

Let AB (Fig. 18) represent the magnitude and direction of 
a force whose line of action is ab, and let M be the origin 
of moments. Let O be the pole, 

and OK { = H) the pole distance a a\b S^ 

of the given force. Draw the \ "^^-^^ \ ^.</ \ 

strings oa, ob, and through 31 k\ — r^O \^--.^o '» 

draw a line parallel to aby inter- \^''' \ ^^""4^. 

secting oa and ob in P and Q. Fig. is \ivi 

Then it is to be proved that the 

moment of the given force with respect to M is equal to 

HxPQ. 

Let // equal the perpendicular distance of M from ab. Then 
the required moment is AB x //. But since the similar triangles 
OAB and RPQ have bases AB, PQ, and altitudes H, h, respec- 
tively, it follows that 

^ = - ; hence, ABxh = PQxH, 
PQ h' 

which proves the proposition. 



36 



GRAPHIC STATICS. 



It should be noticed that PQ represents a length, while H 
represents deforce niagnitiide. Hence, the moment of the given 
force with respect to M is equal to the moment of a force H 
with an arm PQ. [It may, in fact, be shown that the given 
force is equivalent to an equal force acting in the line PQ 
(whose moment about iW is therefore zero), and a couple with 
forces of magnitude H, and arm PQ. For AB acting in ab is 
equivalent to AO and OB acting in ao and ob respectively. 
Also, AO acting in ao is equivalent to forces represented by 
AK and KO acting respectively in PQ and in a line through 
P at right angles to PQ ; and OB may be replaced by forces 
represented by OK and KB, the former acting in a line through 
Q perpendicular to PQ, and the latter in PQ. But AK and 
KB are equivalent to AB ; hence, the proposition is proved.] 

56. Moment of the Resultant of Several Forces. — The mo- 
ment of the resultant of any number of consecutive forces in 
the force and funicular polygons may be found by a method 
similar to that just described. Thus, let Fig. 19 represent the 

force polygon and the 
funicular polygon for six 
forces, and let it be re- 
quired to find the moment 
of the resultant of the four 
forces represented in the 
force polygon by BC, CD, 
DE, and EF, with respect 
to any point M. The re- 
sultant of the four forces 
is represented by BF, and 
acts in a line through the intersection of ob and of. Through 
M draw a line parallel to BF, intersecting ob and of in P and Q 
respectively. Then PQ multiplied by OK, the pole distance of 
BF, gives the required moment. This method does not apply 
to the determination of the moment of the resultant of several 
forces not consecutive in the force polygon. 




SUMMARY OF CONDITIONS OF EQUILIBRIUM. 37 

57. Moments of Parallel Forces. — The method of Arts. 55 
and 56 is especially useful when it is desired to find the mo- 
ments of any or all of a system of parallel forces ; since with 
such a system the pole distance is the same for all forces, and 
the moments are therefore proportional to the intercepts found 
by the above method. 

Example. — Assume five parallel forces at random; choose 
an origin, and determine their separate moments, also the 
moment of their resultant, by the method of Arts. 55 and 56. 

§ 6. Summary of Conditions of Eqtnlibrinm. 

58. Graphical and Analytical Conditions of Equilibrium Com- 
pared. — It has been shown (Art. 35) that the conditions of 
equilibrium for a system of complanar forces acting on a rigid 
body are two in number : 

(1) The force polygon must close. 

(II) Any funicular polygon must close. 
The analytical conditions * are the following : 
(i) The algebraic sum of the resolved parts of the forces 
in any direction must be zero. 

(2) The algebraic sum of their moments for any origin must 
be zero. 

. The condition (i) is readily seen to be equivalent to (I). 
For if the sides of the force polygon be orthographically 
projected upon any line, their projections will represent in 
magnitude and direction the resolved parts of the several 
forces parallel to the line ; and, if the force polygon is closed, 
the algebraic sum of these projections is zero, whatever the 
direction of the assumed line. (See Art. 22.) It may also 
be seen that condition (II) carries with it (2). For, if every 
funicular polygon closes, the system is equivalent to two equal 
and opposite forces having the same line of action (Arts. 31, 

* See Minchins Statics, Vol. I, p. 114. 



38 



GRAPHIC STATICS. 



35, and 50) ; and the sum of the moments of these two forces 
must be zero. 

A further comparison may be made. The analytical condi- 
tion {2) carries (i) with it ; and similarly the graphical condi- 
tion (II) carries with it the condition (I). That (2) includes 
(i) may be seen as follows : If the sum of the moments is zero 
for one origin M-^, there can be no resultant couple, neither can 
there be a resultant force unless with a line of action passing 
through Ml. If the sum of the moments is zero for two 
origins, Mi and M2, the resultant force, if one exists, must 
act in the line MiM^. If the sum of the moments is zero 
also for a third origin Mz, not on the line M1M2, there can 
be no resultant force. It follows at once that condition (i) 
must hold. 

That (II) includes (I) may be shown as follows : Let A and 
E be the first and last points of the force polygon, and choose 
a pole Oi. Then, if the funicular polygon closes, OiA and 
OiE are parallel. Choose a second pole 6^2, and, if the funic- 
ular polygon again closes, O^A and O^E are parallel. AE 
must then be parallel to 6^16^2, unless A and E coincide. 
Now, choose a third pole O^, not on OiO^ ; if the funicular 
polygon for this pole closes, O^A and O^E must be parallel. 
But this is impossible unless A and E coincide ; that is, unless 
condition (I) holds. 

The last result may be reached in another way. With any 
pole O draw a funicular polygon and suppose it to close. The 
system is thus reduced to two forces acting in the same line 
oa. Hence, there is no resultant couple, and if there is a 
resultant force, its line of action is oa. With the same pole 
draw a second funicular polygon, the first side being o^a\ 
parallel to oa. If this polygon closes, there can be no result- 
ant force, for if one existed it would act in the line o'a' ; and 
there would thus be two resultant forces, acting in different 
lines oa and o^a\ which is impossible. 



SUMMARY OF CONDITIONS OF EQUILIBRIUM. 3^ 

59. Summary. — It is now evident that the conditions neces- 
sary to insure equihbrium may be stated in several different 
ways, both analytically and graphically. To summarize : 

A. Analytically : There will be equilibrium if either of the 
following conditions is satisfied : 

(i) The sum of the moments is zero for each of three points 
not in the same line. 

(2) The sum of the moments is zero for each of two points, 
and the sum of the resolved parts is zero for a line not per- 
pendicular to the line joining those two points. 

(3) The sum of the moments is zero for one point, and the 
sum of the resolved parts is zero for each of two directions. 

B. Graphically : There will be equilibrium if either of these 
three conditions is satisfied : 

(i) A funicular polygon closes for each of three poles not in 
the same line. 

(2) Two funicular polygons close for the same pole. 

(3) One funicular polygon closes and the force polygon closes. 



CHAPTER III. INTERNAL FORCES AND 

STRESSES. 

§ I . External and Internal Forces. 

60. Definitions. — It was stated in Art. 3 that every force 
acting upon any body is exerted by some other body. In what 
precedes, we have been concerned only with the effects pro- 
duced by forces upon the bodies to which they are apphed. 
It has therefore not been needful to consider the bodies which 
exert the forces. It is now necessary to consider forces in 
another aspect. 

The forces applied to any particle of a body may be either 
external or internal. 

An external force is one exerted upon the body in question 
by some other body. 

An internal force is one exerted upon one portion of the body 
by another portio7t of the same body. 

It is important to note, however, that the same force may be 
internal from one point of view, and external from another. 
Thus, if a given body be conceived as made up of two parts, 
X and F, a force exerted tcpoit X by F is internal as regards 
the whole body, but external as regards the part X. Thus, let 

AB (Fig. 20) represent 

i f 1^ ^ a bar, acted upon by two 

forces of equal magnitude 
applied at the ends paral- 
lel to the length of the bar, in such a way as to tend to pull it 
apart. These two forces are exerted upon AB by some other 

40 



C B 

Fig. SO 



EXTERNAL AND INTERNAL FORCES. 4 1 

bodies not specified. If the whole bar be considered, the 
external forces acting upon it are simply the two forces 
named. 

But suppose the body under consideration is AC, a portion of 
AB. The external forces acting upon this body are (i) a force 
at A, already mentioned, and (2) a force at C, exerted upon AC 
by CB. This latter force is internal to the bar AB, but external 
lo AC 

61. Conditions of Equilibrium Apply to External Forces. — In 

applying the conditions of equilibrium deduced in previous 
articles, it must be remembered that only external forces are 
referred to. It is also important to notice that the principles 
apply to any body or any portion of a body in equilibrium ; and 
the system of forces in every case must include all forces that 
are external to the body or portion of a body in question. 

Thus, if the bar AB (Fig. 20) be in equilibrium under the 
action of two opposite forces P and Q, applied at A and B 
respectively, as shown in the figure, the principles of equilibrium 
may be applied either to the whole bar, or to any part of it, as 
AC 

{a) For the equilibrium of the whole bar AB, we must have 
P=Q, these being supposed the only external forces acting on 
the bar. 

ib) For the equilibrium of AC, the force exerted upon AC by 
CB must be equal and opposite to P. This latter force is 
external to AC, though internal \.o the whole bar. 

The method just illustrated is of frequent use in the investi- 
gation of engineering structures. It is often desired to deter- 
mine the internal forces acting in the members of a structure, 
and the general method is this : Direct the attention to such a 
portion of the whole structure or body considered that the 
internal forces which it is desired to determine shall be external 
to the portion in question. (See Art. 67.) 



42 



GRAPHIC STATICS. 



2. External and Internal Sti^esses. 



62. Newton's Third Law. — Let X and Y be any two portions 
of matter ; then if X acts upon Y with a certain force, F acts 
upon X with a force of equal magnitude in the opposite direc- 
tion. This is the principle stated in Newton's third law of 
motion, — that ''to every action there is an equal and contrary 
reaction." It is justified by universal experience. 

63. Stress. — Definition. — Two forces exerted by two por- 
tions of matter upon each other in such a way as to constitute 
an action and its reaction, make up a stress. 

Illnstrations. — The earth attracts the moon with a certain 
force, and the moon attracts the earth with an equal and opposite 
force.- The two forces constitute a stress. 

Two electrified bodies attract (or repel) each other with equal 
and opposite forces. These two forces constitute a stress. 

Any two bodies in contact exert upon each other equal and 
opposite pressures (forces), constituting a stress. 

By the magnitude of a stress is meant the magnitude of either 
of its forces. 

64. External and Internal Stresses. — It has been seen that 
two portions of matter are concerned in every stress. Now the 
two portions may be regarded either as separate bodies, or as 
parts of a body or system of bodies which include both. 

A stress acting between two parts of the same body (or 
system of bodies) is an internal stress as regards that body 
or system. 

A stress acting between two distinct bodies is an external 
stress as regards either body. 

It is important to notice that the same stress may be internal 
from one point of view, and external from another. Thus, if 
a given body be considered as made up of two parts X and F, 
a stress exerted between Xand Fis internal X.Q the whole body, 
but external to either X or F 




EXTERNAL AND EXTERNAL STRESSES. 43 

Illustration. — Consider a body AB (Fig. 21) resting upon a 
second body F, and supporting another body X, as shown. If 
the weight of the body AB be disregarded, 
the forces acting upon it are (i) the down- 
ward pressure (say P) exerted by X at the 
surface A, and the upward pressure (say Q) 
exerted by Fat ^ ; these forces being equal 
and opposite, since the body is in equiUbrium. 
Now the body X is acted upon by AB with 
a force equal and opposite to P, and these '^^s- si 

two forces constitute a stress which is external to AB. There 
is also an external stress exerted between AB and Y at B. 
But let AB be considered as made up of two parts, AC and CB. 
Then (Art. 60) CB exerts upon AC 2. force upward, and AC 
exerts upon CB a force downward. These two forces are an 
action and its reaction, and constitute a stress which is internal 
to the body AB. This same stress is, however, external to 
either AC ox CB. An equivalent stress evidently exists at 
every section between A and B. (When we refer to the force 
acting upon CB at C^ we mean the resultant of all forces exerted 
upon the particles of CB by the particles of AC. This resultant 
is made up of very many forces acting between the particles. 
Also the stress at C means the stress made up of the two 
resultants of the forces exerted \>y AC and CB upon each other.) 

65. Three Kinds of Internal Stress. — It is evident that the 
internal stress at C in the body AB (Fig. 20) depends upon 
the external forces applied to the body. If the forces at A and 
B cease to act, the forces exerted hy AC and CB upon each 
other become zero. If the forces at A and B are reversed in 
direction, so also are those at C (As a matter of fact, the 
particles oi AC exert forces upon those of CB, even if the 
external forces do not act. But if the external forces applied 
to CB are balanced, the resultant of the forces exerted on CB 
by ^4(7 is zero.) 



44 



GRAPHIC STATICS. 



The nature of the internal stress at any point in a body is 
thus seen to depend upon the external forces applied to the body. 

Now, if we consider two adjacent portions of a body (as 
the parts X and Y, Fig. 22) separated by a plane surface, the 

external forces may have either of three 

\\~7. \\ tendencies : (i) to pull X and Y apart 

in a direction perpendicular to the plane 
^^' of separation ; (2) to push them together 

in a similar direction ; (3) to slide each over the other along the 
plane of separation. Corresponding to these three tendencies, 
the stress between X and Y may be of either of three kinds : 
tensile^ compressive^ or shearing. 

A tensile stress is such as comes into action to resist a 
tendency of the two portions of the body to be pulled apart in 
the direction of the normal to their surface of separation. 

A compressive stress is such as comes into action to resist a 
tendency of X and Y to move toward each other along the 
normal to the surface. 

A shearing stress is such as acts to resist a tendency of X 
and Fto slide over each other along the surface between them. 

In case of a tensile stress, the force exerted by X upon Y 
has the direction from Y toward X\ and the force exerted by 
Fupon X has the direction from X toward Y. 

In case of a compressive stress, the force exerted by X upon 
Y has the direction from X toward F; and the force exerted by 
Fupon Xhas the direction from F toward X. 

In the case of a shearing stress, the force exerted by X 
upon Fmay have any direction in the plane of separation ; the 
force exerted by Fupon X having the opposite direction. 

If X and F are separate bodies, instead of parts of one 
body, a similar classification may be made of the kinds of 
stress between them ; but with these we shall have no occasion 
to deal. The terms tensile stress, compressive stress, and shear- 
ing stress (or tension, compression, and shea?) are usually applied 
only to internal stresses. 



EXTERNAL AND INTERNAL STRESSES. 



45 



66. Strain. — In what has preceded, the bodies dealt with 
have been regarded as rigid ; that is, the relative positions of 
the particles of any body have been regarded as remaining 
unchanged. But, as remarked heretofore, no known body is 
perfectly rigid. If no external forces act upon a body, its 
particles take certain positions relative to each other, and the 
body has what is called its natural shape and size. If external 
forces are applied, the shape and size will generally be changed ; 
the body is then said to be in a state of strain. The deforma- 
tion produced by any system of applied forces is called the 
strain due to those forces. The nature of this strain in any 
case depends upon the way in which the forces are applied. It 
is unnecessary to treat this subject further at this point, since 
we shall at present be concerned only with problems in the 
treatment of which it will be sufficiently correct to regard the 
bodies as rigid. 

[Note. — There is a lack of uniformity among writers in regard to the meanings 
attached to the words stress and strain. It may, therefore, be well to explain again 
at this point the way in which these words are used in the following pages. The 
word stress should be employed only in the sense above defined, as consisting of two 
equal and opposite forces constituting an action and its reaction. The two forces are 
exerted respectively by two bodies or portions of matter upon each other. An 
internal stress is a stress between two parts of the same body. An internal force is 
one of the forces of an internal stress. It is intended in what follows to use the 
word? "internal stress" (or simply "stress '') only when both the constituent forces 
are referred to; and when only one of the forces is meant, to. use the words "inter- 
nal force" (or simply "force"). It will be noticed, therefore, that in the following 
pages the words '■^ force in a bar " are frequently used where many writers would say 
" stress." This departure from the usage of many high authorities seems justified by 
the following considerations : (i) It agrees with the usage which is being adopted 
by the highest authorities in pure mechanics. (2) It is desirable that the nomencla- 
ture of technical mechanics shall agree with that of pure mechanics, so far as they 
deal with the same conceptions. The definition of strain above given is in conform- 
ity with the usage of the majority of the more recent text-books. But it is not rare 
to find in technical literature the word strain used in the sense of internal stress as 
above defined. Such use of the word should be avoided.] 



46 GRAPHIC STATICS. 

§ 3. Determination of Internal Stresses. 

6'j. General Method. — The stresses exerted between the 
parts of a body may or may not be completely determinate by 
means of the principles already deduced. But in all cases 
these principles suffice for their partial determination. The 
general method employed is always the same, and will now be 
illustrated. As heretofore we deal only with complanar forces. 
Let XY (Fig. 23) represent a body in equilibrium under the 
action of any known external forces as shown. Now conceive 
the body to be divided into two parts as X and Y, separated 
by any surface. The particles of X near the surface exert 
upon those of Y certain forces, and are in return acted 

upon by forces exerted by 
the particles of F. These 
forces are internal as re- 
gards the whole body. In 
order to determine them so 
far as possible, we proceed 
as follows : Let the resultant of all the forces exerted by Y 
upon XhQ called T; then T^is either a single force or a couple. 
Now apply the conditions of equilibrium to the body X. The 
external forces acting on X are Pi, P.2, P^, and T. Since Pj, P^, 
and Pz are supposed known, T can be determined. In fact, 
T is equal and opposite to the resultant of P^, P^, and P^. 

So much can always be determined. But T is the resultant 
of a great number of forces acting on the various particles 
of X \ and these separate forces cannot in general be deter- 
mined by methods which lie within the scope of this work. 

The general principle just illustrated may be stated as 
follows : 

If a body in equilibrium tinder any external forces be conceived 
as made up of two parts X and Y, then the internal forces 
exerted by X upon Y, together with the external forces acting on 
Y, form a system in eqtdlibrinm. 




Fig. S3 



DETERMINATION OF INTERNAL STRESSES. 



47 



As an immediate consequence, we may state that tJie result- 
ant of the forces exerted by X upon V is equivalent to the result- 
ant of the external forces acting on X ; and is equal and opposite 
to the resultant of the external forces acting upon Y. 

Example. — Assume a bar of known dimensions, and the 
magnitudes, directions, and points of application of five forces 
acting on it. Then (i) determine a sixth force which will 
produce equilibrium ; and (2) assume the bar divided into two 
parts and find the resultant of the forces exerted by each part 
on the other. 

6^. Jointed Frame. — In certain ideal cases (corresponding 
more or less closely to actual cases), the internal forces may be 
more completely determined. The most important of these 
cases is that which will be now considered. Conceive a rigid 
body made up of straight rigid bars hinged together at the ends. 
Assume the following conditions : 

(i) The hinges are without friction. 

(2) All external forces acting on the body are applied at 
points where the bars are joined together. 

The meaning of these conditions will be seen by reference to 
Fig. 24. The three bars X, F, and Z are connected by a '* pin 
joint," the end of each bar having a hole 
or ''eye" into which is fitted a pin. 
(Of course the three bars cannot be in 
the same plane, but they may be nearly 
so, and will be so assumed in what fol- 
lows.) Condition (i) is satisfied if the ie-3_4 
pin is assumed frictionless. The effect of this is that the force 
exerted upon the pin by any bar (and the equal and opposite 
reaction exerted upon the bar by the pin) acts in the normal 
to the surfaces of these bodies at the point of contact ; and, 
therefore, through the centers of the pin and the hole. Condi- 
tion (2) means that any external force (that is, external to the 




48 



GRAPHIC STATICS. 



whole body) applied to any bar is applied to the end and in a 
line through the center of the pin. 

With the connection as shown, the bars do not exert forces 
upon each other directly. But each exerts a force upon the 
pin, and any force exerted by V or Z upon the pin causes an 
equal force to be exerted upon X. (This is seen by applying 
the condition of equilibrium to the pin.) Hence, in considering 
the forces acting upon any bar as X, we may disregard the pin 
and assume that each of the other bars acts directly upon X. 
By what has been sai'd, all such forces exerted upon X by the 
other bars meeting it at the joint may be regarded as acting at 
the same point — the center of the pin. We therefore treat the 
bars as mere '^material lines," and regard all forces exerted on 
any bar (whether by the other bars or by outside bodies) as 
applied at the ends of this "material line." 

Since, with these assumptions, all forces acting on any bar 
MN (Fig. 25) are applied either at M or JV, the forces applied 
at M must balance those applied at N. The resultants of the 
two sets must therefore be equal and opposite, and have the 
same line of action — namely MN. Further, it follows that 
the stress in the bar, acting across any plane perpendicular to 
its length, is a direct tension or compression. 



69. Internal Stresses in a Jointed Frame. — Let Fig. 25 rep- 
resent a jointed frame such as above described, in equilibrium 

under any known external forces. 
Let us apply the general method 
of Art. 6y to this case. Divide 
the body into two parts, X and V, 
by the surface AB as shown. Now 
apply the conditions of equilibrium 
to the body X. The system of 
forces acting upon this body consists of Pi, P2, Pq, and the forces 
exerted by Y upon X in the three members cut by the surface 
AB. By Art. 68 the lines of action of these forces are known, 




Fig. 2B 



DETERMINATION OF INTERNAL STRESSES. 49 

being coincident with the axes of the members cut. Hence, 
the system in equilibrium consists of six forces, three com- 
pletely known, and three known only in lines of action. 
The determination of the unknown forces in magnitude and 
direction is then a case under the general problem discussed in 
Art. 40. 

Nature of the stresses. — As soon as the direction of the 
force acting upon X in any one of the members cut is known, 
the nature of the stress in that member (whether tension or 
compression) is known. For a force toivard X denotes com- 
pression ; while a force aivay from X denotes tension. (Art. 

65.) 

If a section can be taken cutting only two members, the 
forces in these may be found by the force polygon alone. The 
same is true, if any number of members are cut, but the stresses 
in all but two are known. 

The methods described in the last three articles will find 
frequent application in the chapters on roof and bridge trusses, 
Part II. 

Example. — Assume a jointed frame similar to the one shown 
in Fig. 25, and let external forces act at all the joints. Then 
(i) assume all but three of the forces known in magnitude and 
direction and determine the remaining three so as to produce 
equilibrium. (2) Take a section cutting three members and 
determine the stresses in those members. 

70. Indeterminate Cases. — If, in dividing the frame, more 
than three members are cut, the number of unknown forces is 
too great to admit of the determination of their magnitudes. 
In such a case, it may happen that a section elsewhere through 
the body will cut but three members ; and that after the deter- 
mination of the stresses in these three, another section can be 
taken cutting but three members whose stresses are unknown. 
So long as this can be continued, the determination of the 



50 



GRAPHIC STATICS. 




internal stresses can proceed. Thus, in Fig. 26, if a section 
be first taken at AB, there are four unknown forces to be 
determined. But, if the section A' B' be first taken, the 
stresses in the three members cut may be determined ; after 

which the section AB will 
\A' introduce but three unknown 

stresses. 

There may, however, be 

cases in which the stresses 
°. ° cannot all be determmed by 

any method. With such ac- 
tually indeterminate cases we shall not usually have to deal. It 
should be noticed, also, that even when only three members 
are cut, the problem is indeterminate if these three intersect in 
a point. As in the case just discussed, this indeterminateness 
may be either actual or only apparent ; in the latter case it may 
be treated as above indicated. 

No attempt is here made to develop all methods that are 
applicable or useful in the determination of stresses in jointed 
frames. Some of these are best explained in connection with 
the actual problems giving rise to them. We have sought here 
only to explain and clearly illustrate general principles. 



71. Funicular Polygon Considered as Jointed Frame. — Let 

ab^ be, cd, da (Fig. 27) be the lines of action of four forces in 
equihbrium, the force polygon being ABCDA. Choosing a 




Fig. 3-7 




pole, draw any funicular polygon, as the one shown. Now let 
the body upon which the forces act be replaced by a jointed 



DETERMINATION OF INTERNAL STRESSES. 



51 



frame whose bars coincide with the sides of the funicular 
polygon. If at the joints of this frame the given forces be 
applied, the frame will be in equilibrium ; and each bar will 
sustain a tension or compression whose magnitude is repre- 
sented by the corresponding ray of the force diagram. 

To prove this, we apply the ''general method" of Art. 6^. 
Consider any joint (as the intersection of oa and ob)^ and let 
the frame be divided by a plane cutting these two members. 
Then the portion of the frame about the joint is acted upon 
by three forces : AB, acting in the line ab, and forces acting in 
the bars cut, their lines of action being oa, ob. If the bar 
oa sustains a compression and ob a tension, their magnitudes 
being represented by OA and BO respectively, the portion 
of the frame about the joint will be in equilibrium. Hence, 
the tendency of the force AB is to produce the stresses men- 
tioned in the bars oa, ob. In the same way it may be shown 
that the tendency of the force BC is to produce in ob and 
oc tensile stresses of magnitudes OB and CO, respectively. 
Applying the same reasoning to each joint, it is seen that every 
part of the frame will be in equilibrium if the bars sustain 
stresses as follows : The bar oa must sustain a compression 
OA ; ob a tension 0B\ oc a tension 0C\ and od 2, compression 
OD. Hence, if the bars are able to sustain these stresses, the 
frame will be in equilibrium. 

If the stress in any member of the frame is a tension, that 
member may be replaced by a flexible string. This is the 
origin of the name string as applied to the sides of the funic- 
ular polygon. This name is retained for convenience, but, as 
just shown, it is not always appropriate. 

72. Outline of Subject. — The foregoing pages, embracing 
Part I, have been devoted to a development of the principles 
of pure statics. We pass next to the application of these 
principles to special classes of problems. 

Part II treats of the determination of internal stresses in 



52 



GRAPHIC STATICS. 



engineering structures. Only "simple" structures are consid- 
ered, — that is, those whose discussion does not involve the 
theory of elasticity. The structures considered include roof 
trusses, beams, and bridge trusses. 

Part III develops the graphic methods of determining cen- 
troids (centers of gravity) and moments of inertia of plane 
areas, including a short discussion of *' inertia-curves." 



Part II. 
STJiESSES IN SIMPLE STRUCTURES. 



-0'.:«<C 



CHAPTER IV. INTRODUCTORY. 

§ I. Outline of Principles and MetJiods. 

73. The Problem of Design. — When any structure is sub- 
jected to the actior. of external forces, there are brought into 
action certain internal stresses in the several parts of the struc- 
ture. The nature and magnitudes of these stresses depend upon 
the external forces acting (Art. 65). In designing the structure, 
each part must be so proportioned that the stresses induced in 
it will not become such as to break or injure the material. 

To determine these internal stresses, when the external forces 
are wholly or partly given, is the problem of design, so far as 
it will be here treated. 

74. External Forces. — The external forces acting on a struc- 
ture must generally be completely known before the internal 
stresses can be determined. These external forces are usually 
only partly given, and the first thing necessary is to determine 
them fully. 

The external forces include (i) the loads which the structure 
is built to sustain, and (2) the reactions exerted by other bodies 
upon the structure at the points where it is supported. The 
former are known or assumed at the outset and the latter are 
to be determined. 

53 



54 



. GRAPHIC STATICS. 



75. Two Classes of Structures. — Structures may be divided 
into two classes, according as they may or may not be treated 
as rigid bodies in determining the reactions. This may be 
illustrated as follows : 

Let a bar AB (Fig. 28) be supported in a horizontal position 
at two points A and B, the supports being smooth so that the 
IP pressures on the bar at 

A and B are vertical. 
Let a known load P be 



A 



1: 



B 



1 



Fij 



if 



applied to the beam at a 
given point, and let it be required to determine the reactions at 
A and B. 

This is a determinate problem ; for there are three parallel 
forces in equilibrium, two being known only in line of action. 
This problem was solved in Art. 38. 

But let AC (Fig. 29) be a rigid bar supported at three points, 
A, B, and C, the reactions at those points being vertical. 

Let any known loads 
be applied at given 
points, and let it be 
required to determine 
the three reactions. 




iiJ 



a 




Fig. SQ 

This problem is indeterminate ; for any number of sets of 
values of the three reactions may be found, which, with the 
applied loads, would produce equilibrium if acting on a rigid 
body. (See Art. 40.) 

Since, however, an actual bar is not a perfectly rigid body, 
such a problem as the one just stated is, in reality, determinate. 
But it cannot be solved without making use of the elastic 
properties of the material of which the body is composed. 

The two classes of problems are, therefore, the following : 
(i) those in which the reactions can be determined by treating 
the structure as a rigid body, and (2) those in which the 
determination of the reactions involves the theory of elasticity. 
We shall at present deal only with the former class of prob- 



OUTLINE OF PRINCIPLES AND METHODS. 55 

lems. Structures coming under this class will be called simple 
structures. 

J^. Truss. — A truss is a structure made up of straight bars 
with ends joined together in such a manner that the whole acts 
as a single body. The ends of the bars are, in practice, joined 
in various ways ; but in determining the internal stresses, the 
connections are assumed to be such that no resistance is offered 
by a joint to the rotation of any member about it. Such a 
structure to be indeformible must be made up of triangular 
elements; for more than three bars hinged together in the 
form of a polygon cannot constitute a rigid whole. If the 
external forces are applied to the truss only at the points where 
the bars are joined, the internal stress at any section of a mem- 
ber will be a simple tension or compression, directed parallel to 
the length of the bar. (See Art. 6'^) 

The most important classes of trusses are roof trusses and 
bridge trusses. The methods used in discussing these classes 
are, of course, applicable to any framed structures under similar 
conditions. 

jy. Loads on a Truss. — The loads sustained by a truss may 
be either fixed or moving. A fixed (or dead) load is one whose 
point of application and direction remain constant. A moving 
(or live) load is one whose point of application passes through 
a series of positions. Fixed loads may be either pennanent or 
temporaiy. 

The loads on a roof truss are usually all fixed, but are of 
various kinds, viz., the weight of the truss itself and of the roof 
covering, which is a permanent load ; the weight of snow 
lodging on the roof, and the pressure of wind, both of which 
are temporary loads. 

A bridge truss supports both fixed and moving loads. The 
former include the weight of the truss itself, of the roadway, 
of all lateral and auxiliary bracing (permanent loads) ; and of 
snow (a temporary load). The latter consist of moving trains 



56 GRAPHIC STATICS. 

in the case of railway bridges, and of teams or crowds of animals 
or people in the case of highway bridges. 

78. Combination of Stresses Due to Different Causes. — When 
a truss is subject to a variety of external loads, it is often 
convenient to consider the effect of a part of them separately. 
If tensile and compressive stresses are distinguished by signs 
plus and minus, the stress in any member due to the combined 
action of any number of loads is equal to the algebraic sum of 
the stresses due to the loads acting separately. 

A proof of this proposition might be given ; but it may be 
accepted as sufficiently evident without formal demonstration. 

79. Beams. — Another class of bodies to be treated .is 
included under the name beam. 

A beam may be defined as a bar (usually straight) resting on 
supports and carrying loads. The loads and reactions are 
commonly applied in a direction transverse to the length of the 
bar ; but this is not necessarily the case.. 

The internal stresses in any section of a beam are less 
simple than those in the bars of an ideal jointed frame such as 
a truss is assumed to be. A discussion of beams is given in 
Chap. VI. 

80. Summary of Principles Needed. — It will be well to 
summarize at this point the main principles and methods which 
will be employed in the discussion of the problems that follow. 

The general problem presented by any structure consists 
of two parts : {a) the determination of the unknown external 
forces (or reactions) and {b) the determination of the internal 
stresses. 

{a) In the case of simple structures the unknown reactions 
are usually two in number, and the cases most commonly pre- 
sented are the following : 

I St. — Their lines of action are known and parallel. 

2nd. — The line of action of one and the point of application 
of the other are known. 



OUTLINE OF PRINCIPLES AND METHODS. 



57 



Since all the external forces form a system in equilibrium, 
these two cases fall under the general problems discussed in 
Arts. 38 and 39. 

{b) In the case of a jointed frame or truss, the lines of 
action of all internal forces are known, since they coincide 
with the axes of the truss members. (Art. 68.) In determin- 
ing their magnitudes we may have to deal with the following- 
problems in equilibrium : 

1st. — The system in equilibrium may be completely known, 
except the magnitudes of two forces. 

2nd. — The magnitudes of three forces may be unknown. 

The first case may be solved by simply making the force 
polygon close. (See Art. 35.) 

The second case may be solved by the method of Art. 40, 
which consists in making the force and funicular polygons 
close ; or by the method of Art. 42 ; or by the principle of 
moments (Art. 51). 

The student should be thoroughly familiar with the prob- 
lems and principles here referred to. In the following chapters 
we proceed to their application. 

81. Division of the Subject. — The subject of the design of 
structures, so far as here dealt with, will be treated in three 
divisions. The first relates to framed structures sustaining 
only stationary loads ; the second to beams sustaining both 
fixed and moving loads ; the third to framed structures sustain- 
ing both fixed and moving loads. Among structures of the 
first class, the most important are roof trusses ; hence, these 
are chiefly referred to in the next chapter. For a similar 
reason, the chapter devoted to the third class of structures 
refers principally to bridge trusses. The chapter on beams 
precede^ that on bridge trusses, for the reason that the methods 
used in dealing with a beam under moving loads form a useful 
introduction to those employed in treating certain classes of 
truss problems. 



CHAPTER V. ROOF TRUSSES. —FRAMED STRUC- 
TURES SUSTAINING STATIONARY LOADS. 

§ I. Loads oil Roof Trusses. 

82. Weights of Trusses. — Among the loads to be sustained 
by a roof truss is the weight of the truss itself. Before the 
structure is designed, its weight is unknown. But, since it is 
necessary to know the weight in order that the design may be 
correctly made, the method of procedure must be as follows : 

Make a preliminary estimate of the weight, basing it upon 
knowledge of similar structures ; or, in the absence of such 
knowledge, upon the best judgment available. Then 'design 
the various truss members, compute their weight, and com- 
pare the actual weight of the truss with the assumed weight. 
If the difference is so great as to materially affect the design of 
the truss members, a new estimate of weight must be made, 
and the computations repeated or revised. No more than one 
or two such trials will usually be needed. 

As a guide in making the preliminary estimate of weight, 
the following forrnulas may be used. They are taken from Mer- 
riman's "Roofs and Bridges," being intended to represent 
approximately the data for actual structures, as compiled by 
Ricker in his "Construction of Trussed Roofs." 

Let /=span in feet ; (^ = distance between adjacent trusses in 
feet; IF^ total weight of one truss in pounds. Then for 
wooden trusses 

and for wrought iron trusses 

J-F=|«/(i+3V/). 

58 



LOADS ON ROOF TRUSSES. 



59 



83. Weight of Roof Covering. — The weight of roof covering 
can be correctly estimated beforehand from the known weights 
of the materials. The following data may be employed, in the 
absence of information as to the specific material to be used. 
(See Merriman's '' Roofs and Bridges," p. 4.) The numbers 
denote the weight in pounds per square foot of roof surface. 

Shingling : Tin, i lb. ; wooden shingles, 2 to 3 lbs. ; iron, i 
to 3 lbs. ; slate, 10 lbs. ; tiles, 12 to 25 lbs. 

Sheathing : Boards i in. thick, 3 to 5 lbs. 

Rafters : 1.5 to 3 lbs. 

Purlins : Wood, i to 3 lbs. ; iron, 2 to 4 lbs. 

Total roof covering, from 5 to 35 lbs. per square foot of roof 
surface. 

84. Snow Loads. — The weight of snow that may have to be 
borne will differ in different localities. For different sections 
of the United States the following may be used as the maximum 
snow loads likely to come upon roofs. 

Maximum for northern United States, 30 lbs. per square foot 
of horizontal area covered. 

For latitude of New York or Chicago, 20 lbs. per square foot. 

For central latitudes in the United States, 10 lbs. per square foot. 

The above weights are given in Merriman's *' Roofs and 
Bridges." They are in excess of those used by some Bridge 
and Roof companies. 

85. Wind Pressure Loads. — The intensity of wind pressure 
against any surface depends upon two elements : {a) the velocity 
of the wind, and {b) the angle between the surface and the 
direction of the wind. 

Theory indicates that the intensity of wind pressure upon a 
surface perpendicular to the direction of the wind should be 
proportional to the square of the velocity of the wind relative to 
the surface. As an approximate law this is borne out by 
experiment. If / denotes the pressure per unit area, and v 
the velocity of the wind, the law is expressed by the formula 



6o 



GRAPHIC STATICS. 



p=ikv^. Here k is proportional to the density of air. Its 
numerical value may be taken as 0.0024, if the units of force, 
length, and time are the pound, foot, and second respectively. 

If the wind strikes a surface obliquely, experiment shows 
that the resulting pressure has a direction practically normal to 
the surface. The tangential component is inappreciable, owing 
to the very slight friction between air and any fairly smooth 
surface. The intensity of the normal pressure depends upon 
the angle at which the wind meets the surface. 

For a given velocity of wind let p^ denote the normal pressure 
per unit area, when the direction of wind makes an angle a 
with the surface, and p,, the pressure per unit area due to the 
same wind striking a surface perpendicularly. Then the follow- 
ing formula * has been given : 



/a 



2 sm a 



-Pn. 



I +SU1" a 

It will rarely be necessary to use values of p^ greater than 
50 lbs. per square foot. The following table gives values of 
the coefficient of p^ in the above formula for different values 
of a. The value of p,^ may be taken as from 40 to 50 lbs. per 
square foot. 



a 


2 sin a 
I + sin^ a 


a 


2 sin a 
I + sin^ a 


0° 


1 
0.00 


50° 


0.97 


10^ 


0.34 


60° 


0.99 


20° 


0.61 


70° 


1. 00 


30° 


0.80 


80^ 


1. 00 


40° 


0.91 


90° 


1. 00 



* This formula is given by various writers. It is cited by Langley (" Experiments in 
Aerodynamics," p. 24), who attributes it to Duchemin. Professor Langley's elaborate 
experiments show so close an agreement with the formula that it may be used without hesi- 
tation in estimating the pressure on roofs. 



ROOF TRUSS WITH VERTICAL LOADS. 6i 

§ 2. Roof Truss with Vertical Loads. 

Z6. Notation. — The method of determining internal stresses 
in the case of vertical loading will be explained by reference to 
the form of truss shown in Fig. 30. The method will be seen 
to be independent of the particular form of the truss. 

For designating the truss members and the lines of action of 
external forces a notation will be employed similar to that used 
in previous chapters. Let each of the areas in the truss dia- 
gram be marked with a letter or other symbol as shown in 
Fig. 30 ; then the truss member or force-line separating any 
two areas may be designated b}^ the two symbols belonging to 
those areas. Thus, the lines of action of the external forces 
are ab, be, eel, etc., and the truss members are gh, hb, hi, etc. 
It is to be noticed that the lines representing the truss mem- 
bers represent also the lines of action of forces, — namely, the 
internal forces in the members. The joint, or point at which 
several members meet, may be designated by naming all the 
surrounding letters. Thus, bcih, Jiijg are two such points. 

Zj. Loads and Reactions. — The loads now considered are 
assumed to be applied in a vertical direction, and to act at the 
upper joints of the truss. This assumption as to the points of 
application may in some cases represent very nearly the facts ; 
in other cases the loads will, in reality, be applied partly at 
intermediate points on the truss members. If the latter is the 
case, the load borne upon any member is assumed to be divided 
between the two joints at its ends. In this case the member 
will be subject not only to direct tension or compression, but to 
bending. With the latter we are not here concerned, although 
it must always be considered in designing the member. 

The ends of the truss are supposed to be supported on hori- 
zontal surfaces, and the reaction at each point of support is 
assumed to have a vertical direction. 

If the loading is symmetrical with reference to a vertical line 



62 GRAPHIC STATICS. 

through the middle of the truss, it is evident that each reaction 
is equal to half the total load. If the loading is not symmetri- 
cal, the reactions cannot be determined so simply. They may, 
however, be readily computed by either graphic or algebraic 
methods. Graphically, the problem is identical with that solved 
in Art. 38. The truss is treated as a rigid body, the external 
forces acting upon it being the loads and reactions, which form 
a system of parallel forces in equilibrium. Two of these forces 
(the reactions) are unknown in magnitude, but known in line 
of action. The construction for determining their magnitudes 
is as follows : 

Draw the force polygon ABCDEF for the five loads ; choose 
a pole O, draw rays OA, OB, OC, etc., and draw the funicular 
polygon as shown in Fig. 30. The two polygons are to be 
completed by including the reactions FG, GA, and both poly- 
gons must close. We may draw first o^, the closing line of the 
funicular polygon, and then the ray OG parallel to it, thus 
determining the point G in the force polygon. The reactions 
are now shown in magnitude and direction by FG and GA. 

88. Determination of Internal Stresses. — When the external 
forces are all known, the internal stresses may be found very 
readily. The only principle needed is, that for any system of 
forces in equilibrium, the force polygon must close. The con- 
struction will now be explained. 

Considering any joint of the truss (Fig. 30) as gJiijg, fix the 
attention upon the portion of the truss bounded by the broken 
line in the figure. This portion is a body in equilibrium under 
the action of four forces whose lines of action coincide with the 
axes of the four bars gJi, hi, ij, jg, respectively. These forces 
are internal as regards the truss as a whole, but external to the 
part in question ; each force being one of the pair constituting 
the internal stress at any point of the bar. Such a force acts 
from the joint if the stress in the bar is a tension ; toward it if 
the stress is a compression. (Art. 69.) 



ROOF TRUSS WITH VERTICAL LOADS. 



63 



Since these four forces form a system in equilibrium, their 
force polygon must close. This condition will enable us to fully 
determine the magnitudes of the forces, provided all but two are 
known, since the polygon can then be constructed as in Art. 18. 




We cannot, however, begin with the joint just considered, 
since at first the four forces are all unknown in magnitude. 

If, however, we start with the joint gabJi, the polygon of 
forces can be at once drawn. For, reasoning as above, it is 
seen that the portion of the truss immediately surrounding 
this joint is in equilibrium under the action of four forces : 
the reaction in the line ga, the load in the line ab, and the 
internal forces in the lines bJi^ Jig. Of these forces, two (the 
reaction and the load) are completely known ; and it is neces- 
sary only to draw a polygon of which two sides represent the 
known forces and the other two sides are made parallel to the 
members bJi, Jig. Such a polygon is shown in Fig. 30 ; and 
BH and HG represent in magnitude and direction the forces 
whose lines of action are bJi, Jig. 



64 GRAPHIC STATICS. 

Evidently, BH and HG represent also the magnitudes (Art. 
63) of the internal stresses in the two members bJi and hg. 
The nature of these stresses may be found as follows : Since 
the four forces represented in the polygon GABHG are in 
equilibrium, and since GA, AS are the directions of two of 
them, the directions of the other two must be BH, HG. 
Hence, BH acts toward the joint and HG from it. This 
shows that the stress in bh is a compression, while that in hg is 
a tension. 

Passing now to the joint bcihb, it is seen that of the four 
forces whose lines of action meet there, two are fully known, 
namely, the load BC acting vertically downward and the inter- 
nal force in hb acting toward the joint (since the stress is com- 
pressive), while the remaining two (viz., the internal forces in 
ci and ih) are unknown in magnitude and direction. Since, 
however, the unknown forces are but two in number, the force 
polygon can be completely drawn, and is represented by the 
quadrilateral HBCIH. The directions of the forces are found 
as in the preceding case, and it is seen that the bars ci and ih 
both sustain compressive stresses. 

The process may be continued by passing to the remaining 
joints in succession, in such order that at each there remain to 
be determined not more than two forces. The complete con- 
struction is shown in Fig. 30. 

It is evident that the loads AB and EF might have been 
omitted without changing the stresses in any of the truss 
members. For their omission would leave as the complete 
force polygon for external forces BCDEGB, and the two 
reactions would be GB and EG ; but the force diagram would 
be otherwise unchanged. 

The great convenience of the notation adopted is now seen.* 



*This is known as Bow's notation. The notation adopted in Part I involves the same 
idea, but it is not usually employed in works on Graphic Statics, though possessing very 
evident advantages. It was suggested to the writer by its use in certain of Professor 
Eddy's works. 



ROOF TRUSS WITH VERTICAL LOADS. 



65 



The line representing the stress in any member is designated 
in the force diagram by letters similar to those which designate 
that member in the truss diagram. The latter is evidently a 
space diagrain {Art. 11). The force diagram is often called a 
stress diagram, since it shows the values of the internal stresses 
in the truss members. 

89. Reciprocal Figures. — There are always two ways of com- 
pleting the force polygon when two of the forces are known 
only in lines of action. (See Art. 18.) Either way will give 
correct results, but unless a certain way be chosen, it will 
become necessary to repeat certain lines in the stress diagram. 
Thus, if, in Fig. 30, instead of GABHG we draw GABH'G, 
the lines GH', H'B are not in convenient positions for use in 
the other polygons of which they ought to form sides. The 
lettering of the diagrams will also be complicated. As an aid 
in drawing the lines in the most advantageous positions, it is 
convenient to remember the fundamental property of figures 
related in such a way as the force and space diagrams shown in 
Fig. 30. 

Such figures are said to be I'ecipi'ocal with regard to each 
other. The fundamental property of reciprocal figures is that 
for every set of lines intersecting in a point in either figure, 
there is in the other a set of lines respectively parallel to them 
and forming a closed polygon. 

It is also an aid to remember that the order of the sides in 
any closed polygon in the stress diagram is the same as the 
order of the corresponding lines in the truss diagram, if taken 
consecutively around the joint. This usually enables us at 
once to draw the sides of each force polygon in the proper 
order. 

90. Order of External Forces in Force Polygon. — It will be 
observed that in the case above considered, in constructing the 
force polygon for the loads and reactions, these forces have 
been taken consecutively in the order in which their points of 



66 



GRAPHIC STATICS. 



application occur in the perimeter of the truss. This is a 
necessary precaution in order that the stress diagram and truss 
diagram may be reciprocal figures, so that no line in the former 
need be duplicated. 

This requirement should be especially noticed in such a case 
as that shown in Fig. 31, in which loads are applied at lower as 




c 
I 

D 

J 

K E 
F 

G 






Fig. 31 



well as at upper joints. If the reactions are found by the 
method of Art. 87, without modification, the force polygon will 
not show the external forces in the proper order, since the 
known forces are not applied at consecutive joints of the truss. 
A new polygon should therefore be drawn after the reactions 
have been determined. 

If desirable (as in some cases it may be) to make use of a 
funicular polygon in which the external forces are taken con- 
secutively, this may be drawn after the reactions are deter- 
mined and the new force polygon is drawn. 

If a load be applied at some joint interior to the truss, as at 
M (Fig. 31), then in constructing the stress diagram it should 



STRESSES DUE TO WIND PRESSURE. 6/ 

be assumed to act at iV, where its line of action intersects the 
exterior member of the truss, and the fictitious member MN 
inserted. The stresses in the actual truss members will be 
unaffected by this assumption, and such a device is necessary in 
order that the stress diagram may be the true reciprocal of the 
truss diagram. 

§ 3. Stresses Due to Wind Pressure. 

91. Direction of Reactions Due to Wind Pressure. — Since 
the effective pressure of the wind has the direction normal to 
the surface of the roof (Art. 85), it has a horizontal component 
which must be resisted by the reactions at the supports. 
These cannot, therefore, act vertically, as in the case when 
the loading is vertical. Their actual directions will depend 
upon the manner in which the ends of the truss are sup- 
ported. 

If the ends of the truss are immovable, the directions of the 
reactions cannot be determined, since any one of an infinite 
number of pairs of forces acting at the ends would produce 
equilibrium. (The same would be true of the reactions due to 
vertical loads.) In such a case the usual assumption is one of 
the following : (i) the reactions are assumed parallel to the 
loads ; (2) the resolved parts of the reactions in the horizontal 
direction are assumed equal. 

In the case of trusses of large span it is not unusual to 
support one end of the truss upon rollers so that it is free to 
move horizontally, the other end being hinged, or otherwise 
arranged to prevent both horizontal and vertical motion. This 
allovvs for expansion and contraction with change of tempera- 
ture, as well as for movements due to the small distortions of 
the truss under loads. With this arrangement the reaction at 
the end supported on rollers must be vertical ; and since the 
point of application of the other reaction is known, both can 
be fully determined by the method described in Art. 39. 



68 GRAPHIC STATICS. 

92. Determination of Reactions. — The methods of finding 
reactions will now be explained for the three cases mentioned 
in the preceding article: (i) Assuming both reactions parallel 
to the wind ; (2) assuming the horizontal resolved parts of the 
two reactions equal ; and (3) assuming one reaction vertical. 

(i) The first case needs no explanation, since it is identical 
with that described in Art. 8y, except that the loads and 
reactions have a direction normal to one surface of the roof, 
instead of being vertical. It is to be noticed that this assump- 
tion cannot be made if the roof surface is curved, since the 
lines of action of the forces will not be parallel. But since 
the direction of the resultant of the loads will be known from 
the force polygon, both reactions may be assumed to act par- 
allel to this resultant, and the construction made as before. 

(2) In the second case, let each reaction be replaced by two 
forces acting at the support, one horizontal and the other 
vertical. The two horizontal forces are known as soon as the 
force polygon for the loads is drawn, and the two vertical 
forces may be found as in the preceding case, since their lines 
of action are known. 

In Fig. 32, let al?y be, cd be the lines of action of the wdnd 
forces. Let the right reaction be considered as made up of a 
horizontal component acting in de and a vertical component 
acting in ef\ and let the left reaction be replaced by a vertical 
component acting in fg and a horizontal component acting in 
ga. Draw the force polygon (or "load-line") ABCD. By the 
assumption already made GA and DE are to be equal, and 
their sum is to equal the horizontal resolved part of AD. 
Through the middle point of AD draw a vertical line; its 
intersections with horizontal lines through A and D determine 
the points G and E, so that the two forces GA and DE become 
known. The only remaining unknown forces are the vertical 
forces ^i^and FG. Choose a pole O, draw rays to the points 
G, A, B, C, D, E, and then the corresponding strings. Through 
the points determined by the intersection of og with^, and oe 



STRESSES DUE TO WIND PRESSURE. 



69 



with cf, draw the string of. The corresponding ray drawn 
from O intersects EG in the point F, thus determining EF and 
FG, The reactions are now wholly known ; that at the left 
support being DF, and that at the right support FA. 



'\d.<'y 




a 



(3) For the third case the construction is shown in Fig. 33 
{A) and {B), for the two opposite directions of the wind. The 
method is identical with that employed in Art. 39. Only one 
point of the line of action of the left reaction is known, hence 
this is taken as the point of intersection of the corresponding 
strings of the funicular polygon. One of these strings can be 
drawn at once, since the corresponding ray is known ; and the 
other is known after the remaining strings have been drawn, 
since it must close the polygon. The construction should be 
carefully followed through by the student. 

The funicular polygons for the two directions of the wind are 
distinguished by the use of O and O' to designate the two poles. 



70 



GRAPHIC STATICS. 



In Fig. 33 (A), ABCDEFGHIA is the force polygon for the 
case when the wind is from the right. Notice that the points 
A, B, C, D, coincide. This means that the loads AB, BC, CD, 
are each zero. 




^^. 



'^-' /- 



■::M 



Fio;. 33 P 



DABC 




=>o 



In diagram {B), ABCDEFGHVA is the force polygon for 
the case when the wind is from the left. The points E, F, G, H, 
coincide, because the loads EF, EG, GH, are each zero. 

93. Stress Diagrams for Wind Pressure. — When the loads 
and reactions due to wind pressure are known, the internal 
stresses can be found by drawing a stress diagram, just as in 
the case of vertical loads. The construction involves no new 
principle, and will be readily made by the student. In Figs. 
33 {A) and 33 {B) are shown the diagrams for the two directions 
of the wind. 



MAXIMUM STRESSES. 



71 



The stresses in all members of the truss must be determined 
for each direction of the wind. If the truss is symmetrical 
with respect to a vertical line, as is usually the case, it may be 
that the same stress diagram will apply for both directions of 
wind. This will be so if the reactions are assumed to act as 
in cases (i) and (2) of the preceding article. In the case repre- 
sented in Fig, 33, however, the hinging of one end of the truss 
destroys the symmetry of the two stress diagrams, and both 
must be drawn in full. 

§ 4. MaximtLin Stresses. 

94. General Principles. — For the purpose of designing any 
truss member, it is necessary to know the greatest stresses to 
which it will be subjected under any possible combination of 
loads. 

Stresses are combined in accordance with the principle stated 
in Art. 'j'^, that the resultant stress in a truss member due to 
the combined action of any loads is equal to the algebraic sum 
of the stresses due to their separate action. 

The method will be illustrated by the solution of an example 
with numerical data. 

95. Problem — Numerical Data. — Let it be required to design 
a wrought iron truss of 40 ft. span, of the form shown in Fig. 
34 (PI. I). Let 12 ft. be the distance apart of trusses, and let 
the loads be as follows : 

Weight of truss, to be assumed in accordance with the 
formula of Art. 82 : W—\ al (i+yo ^)- '^^^^ gives W= 1800 lbs. 
Assuming this to be divided equally among the upper panels, 
and that the load for each panel is borne equally by the two 
adjacent joints, the load at each of the joints be, ed, de is 450 
lbs. The loads at the end joints may be neglected, being borne 
directly at the supports. 

Weight of 7'oof — This depends upon the materials used and 
the method of construction, but will be taken as 6 lbs. per sq. ft. 



72 GRAPHIC STATICS. 

of roof area, giving 900 lbs. as the load at each joint. This, 
also, is a permanent load. Total permanent load per joint, 
1350 lbs. 

Weight of S7101V. — Taking this as 1 5 lbs. per horizontal 
square foot, we find 1800 lbs. as the load at each joint. 

Wind pi'essiLvc. — This is computed from the formula 

2 sin a ^ I \ ^ c ^ 

Pa= , . ^ A- Art. 85.) 

I +sm^ a 

For this case we put sina = -|-|=|; ^,^ = 40 lbs. per sq. ft.; 
whence /a = 35 lbs. per sq. ft. (about). This gives upon each 
panel of the roof 5250 lbs. Then with the wind from either 
side, the wind loads on that side would be 2625 lbs., 5250 lbs., 
2625 lbs. respectively. 

96. Stress Diagrams. — We are now ready to construct the 
stress diagrams. 

The truss is shown (PL I) in Fig. 34 {A). Fig. 34 {B) is the 
stress diagram for permanent loads. No diagram for snow 
loads is needed, since it would be exactly similar to that for 
permanent loads. The snow load at any joint being four-thirds 
as great as the permanent load, the stress in any member due 
to snow is four-thirds that due to permanent loads. 

Fig. 34 {C) shows the stress diagram for the case of wind 
blowing from the left. The reactions are assumed to act in 
lines parallel to the loads — that is, normal to the roof. With 
this assumption, no separate diagram is needed for the case of 
wind from the right, since such a diagram would be exactly 
symmetrical to Fig. 34 {C). For example, the stress in the 
member^/; due to the wind blowing from the right is given by 
the line GM 'vn Fig. 34 {C). 

97. Combination of Stresses. — After the stress diagrams are 
completed for the various kinds of loads, the stresses should 
be scaled from the diagrams and entered with proper sign in a 
table, as follows : 



MAXIMUM STRESSES. 



73 



Member. 


Permanent 
.Load. 


Snow. 


Wind R. 


Wind L. 


Max. 


bh 


- 1525 


- 2030 


— 6270 


- 7925 


— 11480 


ci 


- 1255 


— 1670 


— 6270 


-7925 


— 10850 


hi 


- 360 


- 480 





-5250 


— 6090 


ik 


+ 595 


+ 790 


+ 550 


+ 6650 


- 8035 


^h 


+ 1245 


+ 1660 


+ 2600 


+ 8800 


+ "705 


gk 


+ 730 


+ 975 


+ 2300 


+ 2300 


+ 4005 


gin 


+ 1245 


-f 1660 


+ 8800 


+ 2600 


+ 1 1705 


Ik 


+ 595 


-f 790 


+ 6650 


+ 550 


- 8035 


mi 


- 360 


- 480 


-5250 





— 6090 


dl 


- 1255 


— 1670 


-7925 


— 6270 


— 10850 


em 


- 1525 


- 2030 


- 7925 


— 6270 


— 1 1480 



By combining the results, the maximum stress in each mem- 
ber for any possible condition of loading can be determined. 
The possible combinations of loading are the following : Perma- 
nent load alone ; permanent and snow loads ; permanent load, 
and wind from either direction ; permanent and snow loads, and 
wind from either direction. The student will readily under- 
stand the method of combining the separate results. 

The problem here solved relates to a very simple form of 
truss. With some forms there may occur a reversal of stress 
in certain members, under different conditions of loading. 

It is to be noticed that in the table the word maximum is 
used in its numerical sense, and has no reference to the algebraic 
sign of the stress. 



98. Examples. — In Fig. 35 {A) to {F), are shown several 
forms of truss for which the student may draw stress diagrams, 
assuming loads in accordance with the data given in Arts. 82 
to 85. In determining reactions due to wind pressure, the 



74 



GRAPHIC STATICS. 



three assumptions mentioned in Art. 92 should all be used in 
different cases, that the student may become familiar with the 
principle of each. 




Fi^. 35 



§ 5. Cases Apparently I itde terminate. 

99. Failure of Usual Method. — In attempting to construct 
the stress diagram by drawing the force polygon for each joint 
in succession, as in the cases thus far treated, a difficulty is 
met in certain forms of truss. It may happen that after pro- 
ceeding to a certain point it is impossible to select a joint for 
which the force polygon can be com.pletely drawn, the number 
of unknown forces for every joint being greater than two. 

Thus, in the truss shown in Fig. 36, if the stress diagram is 
started in the usual way, beginning at the left support, the force 
polygons for three joints may be constructed without difficulty, 
thus determining the stresses in bl^ Ik, hn, cm, mn, nk. But 
the force polygon for cdqp7imc cannot be constructed, since 
three forces are unknown, — namely, those in dq, qp, pn. And 
at the joint knpsk, the stresses in 7ip, ps, and sk are unknown. 
The problem, therefore, seems at this point to become indeter- 



CASES APPARENTLY INDETERMINATE. 



75 



minate, since either of the two polygons can be completed in 
any number of ways, so far as the known forces determine. It 
can be shown, however, that this ambiguity is only apparent. 
This may be proved as follows : 

Consider the portion of the truss to the left of the broken 
line M'N' . , It is in equilibrium under the action of eight 
forces ; five of these (four loads and a reaction) are known ; 
the remaining three are the forces in er^ rs, and sk. Now, the 
problem of determining these three unknown forces is the same 
as that treated in Art. 40. It was there found to be a deter- 
minate problem, unless the lines of action of the three unknown 
forces intersect in a point or are parallel. 

That the problem is determinate may be seen also from the 
principle of moments (Art. 51). The eight forces mentioned 
being in equilibrium, the sum of their moments is zero for any 
origin in their plane. Let the origin be taken at the point of 
intersection of the lines of action of two of the unknown forces, 
as er, rs. Then from the principle of moments we have (since 
the moments of the two forces named are zero) : Algebraic sum 
of m.oments of loads and reaction to left of section + moment 
of SK=o. The only unknown quantity in this equation is the 
magnitude of SK, which may, therefore, be determined. The 
other unknown forces may be found in a similar manner, 
the origin of moments being in each case chosen so as to 
eliminate two of the three unknown forces. 

The whole problem of drawing the stress diagram is now 
seen to be determinate. For, as soon as the stress in sk is 
known, the force polygon for the joint knpsk contains but two 
unknown sides, and can be drawn at once. No further diffi- 
culty will be met. 

100. Solution of Case of Failure — First Method. — The 
reasoning of the preceding article suggests two methods of 
treating the so-called ambiguous case. These will now be 
described. 



1^ 



GRAPHIC STATICS. 



The first method is to apply the construction of Art. 40, as 
follows : Referring to Fig. 36, consider .the equilibrium of the 
portion of the truss to the left of the line M'N''. The system 
of forces consists of those whose lines of action are ka, ab, be, 
cd, de^ er, rs, sk. Let them be taken in the order named, and 
draw the force polygon for the known forces. (The reaction 
ka is supposed to be already determined.) The known part of 
the force polygon is KABCDE \ the unknown part is to be 
marked ERSK. Choose a pole O and draw rays to K, Ay By 
C, D, and E ; then draw the corresponding strings of the funic- 
ular polygon. Remembering the method of Art. 40, we draw 
first oe, making it pass through the intersection of er and is 




(the reason for this being that // makes the string os also pass 
through that point) ; then draw in succession od, oe, ob, oay ok. 
The string ok intersects sk in a point through which os must be 
drawn. Hence os must join that point with the starting point 
of the polygon. This completes the funicular polygon, except 
the string or, which must pass through the intersection of er 



CASES APPARENTLY INDETERMINATE. 



77 



and rs in some direction not yet known. This string is not 
necessary to the solution of the problem. 

Since os is now known, OS may be drawn parallel to it from 
O ; and by drawing a line from K parallel to the known direc- 
tion of KS, the point 5 is determined, and KS becomes 
known. 

To find ER and RS it is only necessary to draw from 5 a 
line parallel to rs, and from E a line parallel to cr ; their inter- 
section gives the point R, and the force polygon for the system 
of forces considered is complete. The stresses in ei\ rs, and sk 
being now known, the stress diagram may be completely drawn 
by the usual method. 

It is interesting to notice that the method just described 
determines the lines ER, RS, SK, in their proper position in 
the complete stress diagram. The determination of ER and 
RS by this method is not necessary, since the usual method of 
drawing the stress-diagram can be carried out as soon as SK 
is known. But the independent determination of ER and RS 
by the two methods serves as a check on the accuracy of the 
construction. 

The funicular polygon employed in the above construction, 
so far as it belongs to the external forces, may coincide with 
the corresponding part of the funicular polygon used in deter- 
mining the reactions. If this is desired, two points must be 
observed: (i) the string ^r must be the first drawn, and must 
pass through the intersection of er and ri", and (2) the pole 
must be so chosen that ok will not be nearly parallel to ks. 

It should also be noticed that the construction of the funic- 
ular polygon might begin with the string ok, which should then 
be made to pass through the intersection of 7^s and sk. The 
student will be able to carry out this construction without 
difficulty. 

10 1. Solution of Case of Failure — Second Method. — It will 
now be shown how the apparently ambiguous case can be 



78 



GRAPHIC STATICS. 



& 



treated graphically by the principle of moments. Referring 
again to Fig. 36, consider the portion of the truss to the left 
of section JVPN^ It is acted upon by eight forces, of which 
five (the loads and the reaction) are known, and three (whose 
lines of action are er^ rsj sk) are unknown. The algebraic sum 
of the moments of all these forces about any origin must be 
zero. Let the origin be taken at the point of intersection of 
er and rs, so that the moments of the forces acting in these 
lines are both zero ; then the sum of the moments of the five 
known forces, plus the moment of the force acting in the line 
sky must equal zero. Now the sum of the moments of the 
five known forces may be found by the method of Art. 56. 
Through the origin of moments draw a line parallel to the 
resultant of the forces named (that is, a vertical line), and let 
i equal the length intercepted on it by the strings oe, ok. Then 
the required moment is — zH, where H is the pole distance. 
(The minus sign is given in accordance with the convention 
that left-handed rotation shall be positive.) Let P = unknown 
force in line sk, and h its moment-arm. For the purpose of 
computing the moment, assume the stress in sk to be a tension ; 
then the force P acts toward the right and its moment is posi- 
tive, the value being H-P//. 

Hence, Ph — Hi=0', 

or, P = -H. 

h 

From this equation P may be computed. The computation 
may be made graphically as follows : Draw (Fig. 36) a triangle 
WUV, making WU=H (force units) and lVV=k (linear 
units). Lay off lVV=i (linear units) and draw VX parallel 
to VU. Then WX (force units) represents P. This is readily 

seen, since from the two similar triangles we have the propor- 

P i 
tion -- = -, which agrees with the equation above deduced. 

H h 
The computation is simplified if the pole distance H is taken 

equal to as many force units as // is linear units ; or if // 



CASES APPARENTLY INDETERMINATE. 



79 



is some simple multiple of Ji. For, suppose H=7ih\ then 

P = lL'A = ni. If ;/-i, P = i. 
h 

The stress in sk is found to be a tension, since P is positive. 
Whatever the nature of the stress, it may be assumed a ten- 
sion in writing the equation, and the sign of the value found 
will show whether the assumption coincides with the fact. 

I02. Other Methods for Case of Failure. — In certain cases 
the method of treating the '' ambiguous case " may profitably 
be varied. 

(i) The construction of Art. lOO may be modified as fol- 
lows : Determine the resultant of the known external forces 
acting on the portion of the truss to one side of the section 
M'N'. This resultant is in equilibrium with the three unknown 
forces acting in the members er, rs, sk. Hence these forces can 
be determined by the special method explained in Art. 42. 

The resultant of the five known forces is represented in 
magnitude and direction by KE in the force polygon ; and its 
line of action passes through the intersection of the strings ocy 
ok in the funicular polygon. Since this point of intersection is 
likely to be inaccessible, the construction of Art. 42 cannot be 
conveniently applied. It may be modified by using instead of 
KE the two forces KA (the reaction in the line ka) and AE 
(the resultant of the four loads, its line of action being deter- 
mined by the intersection of the strings oa and oe). First 
determine forces in the three lines er, rs, sk, which shall be in 
equilibrium with KA ; then make a similar construction for 
AE, and combine the results. 

(2) It has been proposed to employ the following reasoning : 
Remove the members/^ and qr, and insert another represented 
by the broken line in Fig. 36. Evidently this does not change 
the stress in the member sk, since the forces acting on the 
truss to the left of the section M^N^ are unchanged. But with 
this change the difficulty encountered in constructing the stress 
diagram by the usual method is avoided. For when the joint 




8o GRAPHIC STATICS. 

nmcdqpii is reached, the forces acting there will be all known 
except two. Let the stress diagram be drawn in the usual way 
until the stress in sk is known. Then restore the original 
bracing and repeat the construction, using the value just deter- 
mined for SK. 

This method is convenient whenever it is applicable. Cases 
may, however, arise, in which it will fail. For instance, if 
a load is applied at the joint pqrsp, the members pq and qr 

cannot both be omitted, and 
the method cannot be applied. 
Again, it would be inapplicable 
^'^^' ^^ in case of a truss such as shown 

in Fig. 37, with loads at all upper joints. In such cases, one of 
the methods explained in the preceding articles may be applied. 
Other methods might be mentioned, but the foregoing dis- 
cussion of the case will probably be found sufficient. 

103. Failing Case in Other Forms of Truss. — The usual 
method of constructing the stress diagram may fail in other 
forms of truss, though the one above described is the most 
common. In such a case, the problem of finding the stresses 
may be really indeterminate, or only apparently so. Whenever 
it is possible to divide the truss into two parts by cutting three 
members which are not parallel and do not intersect in one 
point, the stresses in the three members cut are determinate as 
soon as all external forces are known, and can be found by 
methods already given. If more than three members are cut, 
the problem of finding the stresses in them is indeterminate, 
unless all but three of these stresses are known. By remem- 
bering these principles, the determinateness of any given prob- 
lem may readily be tested. (See Art. 70.) 

§ 6. Three-Hinged Arch. 

104. Arched Truss Defined. — If a truss is so supported that 
when sustaining vertical loads the reactions at the supports 



THREE-HINGED ARCH. 



8l 



have horizontal components directed toward the centre of the 
span, the trass becomes an arcJi. The only kind of arch we 
shall here consider is that consisting of two partial trusses 
hinged together at the crown, and each hinged at the point of 
support. 

Such a truss is shown in Fig. 38, in which the two partial 
trusses are hinged to the abutments at P and Q, and connected 
by a hinge at the point R. Since a hinge at the support allows 
the reaction of the supporting body upon the truss to take any 
direction in the plane of the truss, the directions of the reactions 
at P and Q are unknown, as is also that of the force exerted by 
either partial truss upon the other at the point R. The problem 
of determining the reactions may, therefore, at first sight seem 
indeterminate. It will be shown in the next article that it is 
in reality determinate, and that the only principles needed in 
the solution are such as have been already often applied in the 
preceding chapters. The three-hinged arch is indeed a *' simple " 
structure (Art. 75), since the theory of elasticity is not needed 
in the determination of the reactions. 

105. Reactions Due to a Single Load. — The method of find- 
ing the reactions is most clearly understood by considering the 




effect of a single load on either partial truss. Let a load be 
applied at 5 (Fig. 38) and let all other loads acting on either 
portion of the truss (including the weight of the structure) be 



82 GRAPHIC STATICS. 

neglected. Call the two partial trusses X and V, and consider 
the part V. The only forces acting upon it are the reaction 
exerted by the abutment at Q and the force exerted at R by 
the truss X. These two forces, being in equilibrium, must 
have the same line of action, which is, therefore, the line QR. 
Consider now the body X. The external forces acting upon it 
are the load at S, the reaction of the abutment at P, and the 
force exerted by V at the point R. But this last force is equal 
and opposite to the force exerted by X upon F, and its line of 
action is therefore QR. The three forces acting upon the 
body X being in equilibrium, their lines of action must meet in 
a point ; which point is found by prolonging QR to meet the 
line of action of the applied load. Let T be this point, then 
PT is the line of action of the reaction at P. The reactions 
can now be determined by drawing the triangle of forces. This 
triangle is ABC in the figure, AB representing the load at S, BC 
the reaction in the line QR (also marked be), and CA the reac- 
tion in the line PT (marked also cci). Evidently ABC may be 
regarded as the polygon of external forces, either for the partial 
truss X, or for the whole structure composed of X and Y\ and 
BC represents either the force exerted by Y upon X at R, or 
the force exerted upon Y by the supporting body at Q. 

If, now, the structure be loaded at other points, the reactions 
due to each load may be found separately ; the resultant of 
all such separate reactions at either support will be the true 
reaction at that support when all the loads act together. A 
convenient method of applying these principles will be given in 
the next article. 

1 06. Reactions and Stresses Due to Any Vertical Loads. — In 

Fig. 39 is represented a truss consisting of two parts supported 
by hinges at P^ and Q and hinged together at R^ . Consider 
all vertical loads to be applied at the upper joints, their lines of 
action being marked in the usual way. We shall first explain 
the construction for finding the reactions at the supports ; after 



THREE-HINGED ARCH. 



83 



these are determined, the drawing of the stress diagram will 
present no difficulty. 

Since we shall sometimes deal with one of the partial trusses, 
and sometimes with the two considered as a single body, it will 
be well at the outset to specify the external forces acting on 
each of these bodies. 







Fig.SQ 



(i) For the partial truss at the left we have five applied loads, 
the reaction at R' (exerted by the other truss), and the reaction 
at P'. The force polygon for these forces will be marked as 
follows: ABCDEFLA. (The meaning of the letters will be 
understood before the force polygon is actually drawn, by 
reference to the corresponding letters in the space diagram.) 



84 GRAPHIC STATICS. 

(2) For the right-hand partial truss the external forces are 
five loads and two reactions, and the force polygon will be 
marked thus : FGHIJKLF. (Notice that FL and LF are equal 
and opposite forces, being the ''action and reaction" between 
the two trusses at i?'.) 

(3) For the combined structure the external forces are the 
ten loads and the reactions at P^ and Q . (The action and 
reaction at R become now internal forces.) The force polygon 
will be marked thus : ABCDEFGHIJKLA. 

Begin the construction by drawing the force polygon for the 
ten loads on the whole structure, lettering it as just indicated. 
Choose a pole (9, draw the rays, and then the funicular polygon 
as far as possible. Now consider the right partial truss as 
unloaded. The resultant of the remaining five loads is repre- 
sented in magnitude and direction by AF\ its line of action 
must pass through the intersection of oa and of, and is therefore 
the line marked af. Now, reasoning as in the preceding article, 
we see that the reaction at Q must act in the line Q R\ Let 
Q R^ intersect af in T\ then P^ T is the line of action of 
the reaction at P\ Hence the complete force polygon for the 
whole truss, when the right half is unloaded, may be found by 
drawing from F d. line parallel to Q R' , and from A a line parallel 
to P^ T , prolonging them till they intersect at V . The reaction 
at P^ is V A, and that at Q is FV . (The line of action of the 
latter is marked //'.) 

Next, consider the left half to be without loads, the other 
half being loaded. The resultant of the five loads now acting is 
FK, its line of action fk being drawn through the point of 
intersection of of and ok. The reactions at P^ and Q for the 
present case have lines of action P^ R^ and Q' T" , found just 
as in the first case of loading. These reactions are therefore 
determined in magnitude and direction by drawing from A' a 
line parallel to Q' T" and from F a line parallel to P'R[, pro- 
longing them till they intersect at L" . The complete force 
polygon for this case of loading is therefore FGHTJKL"F. 



THREE-HINGED ARCH. 



85 



Consider now that both parts of the truss are loaded. The 
reaction at P^ is the resultant of the two partial reactions V A, 
L"F, and the reaction at Q' is the resultant of the two partial 
reactions FL' , KL". From Z" and L' draw lines parallel 
respectively to FL' and FL" , intersecting in L. Then L" L is 
equal and parallel to FL' ^ and LL' is equal and parallel to L" F. 
Hence KL and LA represent the resultant reactions at Q' and 
P' respectively. This completes the polygon of external forces 
for the whole truss, as well as that for each partial truss. 

The stress diagram can now be drawn in the usual way, 
beginning at the point P' . The diagram for one partial truss 
is shown in Fig. 39 {B). 

If loads are applied at the lower joints of the trusses, the 
reactions due to these may be found in the same manner as for 
the upper loads. But before beginning the determination of 
the stresses, the polygon must be drawn for the external forces 
taken in order around the truss. (See Art. 90.) 

107. Case of Symmetrical Loading. — If the two half trusses 
are exactly similar and symmetrically loaded, the determination 
of reactions and stresses is much simplified. 

(i) As regards the reactions, symmetry shows that the forces 
exerted by the two trusses upon each other at R' are horizontal. 
Hence, referring to Fig, 39, and considering either half-truss, 
as that to the left, the line of action of the reaction at P' may 
be found by drawing a horizontal line through R' and prolong- 
ing it to intersect af, the line of action of the resultant of all 
loads on the left truss ; the line joining this point of intersection 
with P' is the required line of action of the reaction at the left 
abutment. The two reactions are now determined by drawing 
from F 3. horizontal line and from A a line parallel to the line 
just determined, and prolonging them till they intersect. 

(2) As to the stresses, only one partial truss need be consid- 
ered, since the stress diagrams for the two portions will be 
symmetrical figures. 



S6 



GRAPHIC STATICS. 



These principles might have been employed in the case dis- 
cussed in the preceding article ; but the method there used is 
applicable to cases in which neither the trusses nor the loads 
are symmetrical. 

1 08. Wind Pressure Diagram. — The diagrams for wind 
pressure will present no difficulty. The determination of the 
reactions will, indeed, be simpler than in the preceding case, 
since only one partial truss will be loaded at any one time, and 
the line of action of one reaction is therefore known at the 
outset. The construction is shown in Fig. 40. 




Fig. 40. 



In computing wind pressure loads, it will be assumed that 
each joint sustains half the pressure coming upon each of the 
two adjacent panels. It will be sufficiently correct in comput- 
ing the pressure at any joint, to assume the slope of the roof 



THREE-HLNGED ARCH. 



8/ 



as that of the tangent to the roof curve at the joint in ques- 
tion ; and the direction of the wind load may be taken as that 
of the normal to the roof curve at the joint. The force poly- 
gon for the wind loads is therefore not a straight line when the 
roof is curved. In Fig. 40, this polygon is FGHIJK. The 
reactions are to be marked KL, LF. (Since there are no loads 
on the other half-truss, the points ABCDEF will coincide.) 
Choosing a pole O, draw the funicular polygon as shown, and 
determine the line of action of the resultant of all the wind 
loads. This line of action//^ is drawn parallel to FK, through 
the point of intersection of the strings of and ok. Prolong//^ 
to intersect P^R^ produced at 7""; then Q'V is the line of 
action of the reaction at Q'. The force polygon may now be 
completed by drawing KL parallel to QT" and LF parallel to 
P^R\ The reactions KL and LF being thus determined, the 
stress diagram can be drawn without difficulty. 

The stress diagram is drawn for both partial trusses, with the 
result shown in the figure. If the two trusses are symmetrical, 
the diagram for the other direction of the wind need not be 
drawn ; the stresses for this case being found from the diagram 
already drawn. If, however, the partial trusses are dissimilar, a 
second wind-diagram must be drawn. 

It is not necessary to draw separate space diagrams for verti- 
cal loads and wind-forces. The constructions of Figs. 39 and 
40 have been here kept wholly separate, in order that the 
explanation may be more easily followed. 

109. Check by Method of Sections. — In case of a truss of 
long span, especially when the members have many different 
directions and are short compared with the whole length of the 
span, the small errors made in drawing the stress diagram are 
likely to accumulate so much as to vitiate the results. Thus, 
in Fig. 39, if, in drawing the stress diagram, we begin at P' and 
pass from joint to joint, there is no check upon the correctness 
of the work until the point R^ is reached. At that point we 



S8 GRAPHIC STATICS. 

have a second determination of the reaction exerted by each 
half-truss upon the other ; and it is quite hkely that the two 
vahies found will not agree. 

By a method like that employed in Art. lOO for the "inde- 
terminate" form of truss, we may avoid the necessity of mak- 
ing so long a construction before checking the results. In 
Fig. 39 take a section cutting the three members cq, qr, rl, 
and apply the principles of equilibrium to the body at the 
left of the section. The forces acting on this body are LA, 
AB, BC, CQ, QRy RL. Choose a pole O^ and draw the funic- 
ular polygon for this system, making the string o^c pass through 
the point of intersection of cq and qr. Draw successively the 
strings o^c^ o^b, o^a, o'l, prolonging the last to intersect /r. 
Through the point thus determined, draw ^'r, which must also 
pass through the point of intersection of cq and qr. As soon as 
o^r is known, the corresponding ray O^R can be drawn in the 
force diagram, and then by drawing from L a line parallel to 
Ir, the point R is determined. We may now close the force 
polygon, since the directions of the two remaining forces {cq 
and qr) are known. 

The stresses in the three members cut being now known, we 
have a check on the correctness of the construction of the 
stress diagram, as soon as these members are reached in the 
process. 

This method will be found of great use, not only for this 
form of truss, but for any truss containing many members. 

§ 7. Coiinterbracing. 

no. Reversal of Stress. — If the loads supported by a truss 
are fixed in position and unchanging in amount, the stress in 
any member remains constant in magnitude and kind. But in 
most cases such are not the conditions, and it may happen that 
under different combinations of loading, the stress in a certain 
member is sometimes tension and sometimes compression. It 



COUNTERBRACING. 



89 



is often thought desirable to prevent such changes of stress, 
the design of the members and their connections being thereby 
simpHfied. To accomplish this is the object of counterbracing. 

III. Counterbracing. — Consider a truss such as the one 
shown in Fig. 41, subjected to vertical loads and to wind pres- 
sure from either side. A diagonal member such as xy may 
sustain tension under vertical loads alone, or with the wind 
from the left ; while with the wind from the right it may sustain 
compression. 

Now suppose xy removed, and a member represented by the 
broken line xy^ introduced. It may easily be shown that any 




41 



-^.. L 



system of loading which would cause compression in xy will 
cause tension in this new member ; and vice vei'sa. For, divide 
the truss by a section MN cutting xy and two chord mem- 
bers as shown, and let L be the point of intersection of the 
two chord members produced. The kind of stress in xy or 
the other member (whichever is assumed to be present) may 
be determined by considering the system of forces acting on 
one portion of the truss, as that to the left of the section MN. 
Let the principle of moments be applied to this system, the 
origin being taken at Z. If the external forces acting on the 
portion of the truss considered be such as to tend to cause 
right-handed rotation about L, the stress in xy must be com- 
pression in order to resist this tendency ; while, if xy be replaced 
by xy\ a tension must exist in that member to resist right- 
handed rotation about L. Similarly, a tendency of the external 
forces to produce left-handed rotation about L would be resisted 
by a tension in xy ; or by a compression in the member xy\ 
If the two members act at the same time, the stresses in 



90 GRAPHIC STATICS. 

them will be indeterminate. These stresses may, however, be 
made determinate by the following device : 

Let the member xy be so constructed that it cannot sustain 
compression. Then, whenever the external forces are such as 
to tend to throw compression upon it, it ceases to act as a 
truss-member, and the member xy^ receives a tension which is 
determinate. 

If the member xy^ be constructed in the same way, any 
tendency to throw compression upon it causes it immediately 
to cease to act, and puts upon the member xy a tension instead. 

A member such as xy\ constructed in the manner mentioned, 
is called a coiinterbrace. 

112. Determination of Stresses with Counter bracing. — The 

use of counterbracing adds somewhat to the labor of deter- 
mining the maximum stresses, since the members actually under 
stress are not always the same. The method of treating such 
cases will be illustrated in the next article by the solution of an 
example ; but first the main steps in the process may be outlined 
as follows : 

{(i) Construct separate stress diagrams for vertical loads and 
for wind in each direction, assuming the diagonals in all panels 
to slope the same way. 

ip) Determine by comparison of these diagrams in which of 
the diagonal members the resultant stress is ever liable to be a 
compression. Draw in counters to all such members. 

if) With these counterbraces substituted for the original 
members, either draw new stress diagrams, or make the neces- 
sary additions to those already drawn. If the latter method is 
adopted, the added lines should be inked in a different color 
from that employed originally. (In some cases, this construc- 
tion may be unnecessary on account of symmetry, as will be 
illustrated in the next article.) 

{d) Combine the separate stresses for maxima in the usual 
way. 



COUNTERBRACING. 



91 



113. Example. — In Fig.' 42 (PI. II) are given stress diagrams 
for a "bow-string" roof-truss in which counterbracing is em- 
ployed. At (A) is shown the truss or space diagram. The 
span is 48 ft. ; rise of top chord, 16 ft. ; rise of bottom chord, 
8 ft. The chords are arcs of parabolas. The whole truss is 
divided into six panels by equidistant vertical members. The 
distance apart of trusses is taken as 12 ft. 

Assume the weight of the truss at 2400 lbs., and that of the 
roof at 3600 lbs. ; this makes the total permanent load 1000 lbs. 
per panel. Take 800 lbs. as the load at each upper joint, and 
200 lbs. as the load at each lower joint. 

The snow load, computed at 15 lbs. per horizontal square 
foot, is about 1440 lbs. per panel. 

Wind pressure is to be computed from the formula of Art. 85. 

We now proceed to apply the method outlined in the preced- 
ing article. 

(a) Assuming one set of diagonals present, we construct the 
stress diagrams for the various kinds of loads. 

Diagram for permanent loads. — This is shown at {B) Fig. 
42, which needs little explanation. It will be noticed that the 
stress in every diagonal member is zero. This will always be 
the case if the chords are parabolic and the vertical loads are 
equal and spaced at equal horizontal distances. 

Diagram for snow loads. — Fig. 42 {C) is the stress diagram 
for snow loads. In this case, also, the stresses in the diagonals 
are all zero. 

Wind diagrams. — At {D) and {E) are shown the diagrams 
for the two directions of the wind. The only thing needing 
special mention is the method used in laying off the force- 
polygon for the wind loads. We first compute the normal wind 
pressure on each panel by the formula of Art. 85. We thus 
find, when the wind is from the right, the following total pres- 
sures, taking /„ = 40 lbs. per sq. ft. : 

On panel d, 1650 lbs. On panel e, 3870 lbs. On panel/, 5480 lb& 



92 



GRAPHIC STATICS. 



In Fig. 42 {D) these are laid off successively in their proper 
directions to the assumed scale. Thus CD\ U E\ E'G repre- 
sent respectively 1650 lbs. normal to dq, 3870 lbs. normal to es, 
and 5480 lbs. normal to ft. Now each of these loads is to be 
equally divided between the two adjacent joints. Bisect CD' at 
D, UFJ at E, and EG at F\ then CD, DE, EF, EG represent 
the loads at the joints cd, de, ef, and fg respectively. The 
"load line" is therefore CDEFG. 

The reactions are assumed to be parallel to the resultant 
load. With this explanation the figures {D) and {E) will be 
readily understood. 

{b) Comparison of results. — The stresses due to permanent 
loads, wind right, and wind left, are shown in tabular form for 
convenience of comparison. 



Member. 


Perm. Load. 


Snow Load. 


Wind R. 


Wind L. 


Max. 


"J 


- 6660 


- 9680 


— 


4250 


— 


14070 


- 16340 


bl 


-5350 


-7780 


- 


4880 


- 


9650 


- 13130 


C7l 


-4550 


— 6640 


- 


6380 


- 


6200 


— 1 1 190 


dq 


-4550 


— 6640 


- 


9500 


- 


4250 


— 14050* 


es 


-5350 


- 7780 


- 


15030 


- 


3450 


— 20380* 


fl 


-6660 


-9680 


- 


14070 


- 


4300 


— 20730* 


Jh 


+ 5080 


+ 7400 


+ 


780 


+ 


13500 


+ 12480 


kh. 


+ 4680 


+ 6820 


+ 


700 


+ 


12450 


+ 1 1 500 


mh^ 


+ 4470 


+ 6520 


-V 


1950 


+ 


7350 


4- 10990 


Ph 


+ 4470 


+ 6520 


+ 


4100 


+ 


4100 


+ 10990* 


r/z., 


+ 4680 


+ 6820 


+ 


7680 


+ 


2050 


+ 12360* 


th[ 


+ 5080 


+ 7400 


+ 


13500 


+ 


800 


+ 18580* 


jk 


+ 1 180 


+ 1440 


+ 


150 


+ 


2650 


+ 2620 


Im 


+ 1 180 


+ 1440 


— 


280 


+ 


4100 


+ 2620 


lip 


+ 1 180 


+ 1440 


— 


950 


+ 


3800 


+ 2620* 


qr 


+ n8o 


+ 1440 


- 


1650 


+ 


2525 


r + 2620* )^ 
I - 470* i 
r + 2620* ) 


St 


+ 1 180 


+ 1440 




1350 


+ 


1250 


I - 170* i 


kl 








+ 


1300 


- 


4600 


-\- 1 300* 


mn 








+ 


2700 


- 


4100 


4- 2700* 


pq. 








+ 


4850 


- 


3150 


+ 4850* 


rs 








+ 


7080 




1950 


+ 7080* 



COUXTERBRACING. g. 

It is seen that the diagonals shown are all in tension when 
the wind is from the right, and all in compression when the 
wind is from the left. Therefore counters are needed in all 
panels, and the counters will all come in action whenever the 
wind is from the left. 

(c) Stresses in counterbraees. — It is evident from symmetry 
that no new diagrams are needed to determine the stresses in 
the counterbraees. In fact, the counterbrace in each panel is 
situated symmetrically to the main diagonal in another panel, 
and is subject to exactly equal stresses. 

{d) Coinbinatio7i for maxima. — In combining the results for 
the greatest stresses in the various members, it will be assumed 
that the greatest snow load and the greatest wind load can 
never act simultaneously. For each member, therefore, the 
stress due to permanent load is to be added to the snow stress 
or the wind stress, whichever is greater. Again, in combining 
the tabulated results, we consider only the columns headed per- 
manent load, snow load, and wind right ; since whenever the 
wind is from the left, the other system of diagonals is in action. 
This gives the results entered in the last column. 

We now notice the following facts : 

(i) The results given for the diagonal members are the true 
maximum stresses. 

(2) The stress found for each diagonal applies also to the 
symmetrically situated counterbrace. 

(3) For any other member, we are to choose between the 
maximum found for that member and the value found for the 
symmetrically situated member. Thus, —20730 is the true 
maximum stress for both <^y and // ; etc. (The numbers denot- 
ing true maximum stresses are marked with a (*) in the table.) 

It is seen that the verticals, with one exception, may sustain 
a reversal of stress. Thus, 7"/^ and st must be designed for a 
tension of 2620 lbs., and also for a compression of 170 lbs. ; 
while Im and qr are each liable to 2620 lbs. tension and to 470 
lbs. compression. 



CHAPTER VI. SIMPLE BEAMS. 
§ I. Genei'al Principles. 

1 14. Classification of Beams. — A beam has been defined in 
Art. 79. Beams may be treated in two main classes, the basis 
of classification being that described in Art. 75. These two 
classes will be called simple and non-simple beams respectively. 
The present chapter deals only with simple beams, the defini- 
tion of which may be stated as follows : 

A simple beam is one so supported that it may be regarded 
as a rigid body in determining the reactions. 

A simple beam may rest on two supports at the ends ; or it 
may overhang one or both supports. 

A cantilever is any beam projecting beyond its supports. 
Such a beam may be either simple or non-simple. 

A continuous beam is one resting on more than two supports. 
Such a beam is non-simple. 

A beam may be supported in several ways. It is simply 
supported at a point when it rests against the support so that 
the reaction has a fixed direction. It is constrained at a point 
if so held that the tangent to the axis of the beam at that 
point must maintain a fixed direction. If hinged at a support, 
the reaction may have any direction. We shall deal mainly 
with the case of simple support. 

In what follows it will be assumed that the beam rests in a 
horizontal position, since this is the usual case. 

11$. External Shear, Resisting Shear, and Shearing Stress. 

— The external sJiear at any section of a beam is the algebraic 

94 



GENERAL PRINCIPLES. 



95 



sum of the external vertical forces acting on the portion of the 
beam to the left of the section. 

The resisting sJiear at any section is the algebraic sum of the 
internal vertical forces in the section acting on the portion of 
the beam to the left, and exerted by the portion to the right 
of the section. 

The shearing stress at any section is the stress which con- 
sists of the internal vertical forces in the section, exerted by 
the two portions of the beam upon each other. It consists of 
the resisting shear and the reaction to it. (See Art. 63.) 

Let AB (Fig. 43) be a beam in equilibrium under the action 
of any external forces. At any point in its length, as C, con- 
ceive a plane to be passed perpen- 
dicular to the axis of the beam, and ^ ' — ^ 

consider the portion AC, to the left fi^. 43 

of the section. The principles of equilibrium apply to the body 
AC, and the external forces acting upon it include, besides 
those forces to the left of C that are external to the whole bar, 
certain forces acting across the section at C that are internal to 
AB, but external to AC. (Art. 61.) These latter forces' com- 
prise that constituent of the internal stress between AC and 
CB which is exerted by CB upon AC 

Represent by Fthe algebraic sum of the resolved parts in 
the vertical direction of all forces acting on AB to the left of 
the section at C, upward forces being called positive. V is the 
external shear at the given section as above defined. 

Since the body AC is in equilibrium, condition (i), Art. 58, 
requires that the algebraic sum of the resolved parts in the 
vertical direction of all forces acting on it must equal zero. 
Hence the forces acting on AC in the section at C must have a 
vertical component equal to — /^ This vertical component is 
called the resisting shear in the given section. This resisting 
shear is one of the forces of a stress of which the other is an 
equal and opposite force exerted by ^67 upon CB. This stress 
is called the shearing stress in the section, and is called positive 



g6 GRAPHIC STATICS. 

when it resists a tendency of AC to move upward, and of CB 
to move downward. 



1 1 6. Bending Moment, Resisting Moment, and Stress Moment. 

— The bending moment at any section of a beam is the algebraic 
sum of the moments of all the external forces acting on the 
portion of the beam to the left of the section ; the origin of 
moments being taken in the section. 

The resisting moment at any section is the algebraic sum of 
the moments of the internal forces in the section acting on the 
portion of the beam to the left, and exerted by the portion to 
the right of the section ; the origin of moments being the same 
as for bending moment. 

The stress moment or moment of internal stress at any section 
consists of the two equal and opposite moments of the forces 
exerted across the section by the two portions of the beam 
upon each other. 

Referring again to Fig. 43, let us analyze further the forces in 
the section at C. Applying to the body AC the second condition 
of equilibrium ((2) of Art. 58), and taking an origin at a point 
in the section, we see that the algebraic sum of the moments 
of all the external forces acting on the beam to the left of the 
section plus the sum of the moments of the internal forces act- 
ing on ^6^ in the section must equal zero. The former sum is 
defined as the bending moment at the given section. Represent 
it by M. The latter sum is defined as the resisting mome7it at 
the section, and must be equal to —M, by the above principle. 

We have thus far referred only to the internal forces exerted 
by CB upon AC\ but evidently the equal and opposite forces 
exerted by ^4 (7 upon CB have a moment numerically equal to 
M. The equal and opposite moments of the equal and opposite 
forces of the stress in the section together constitute the stress 
moment in the section. 

If the external forces applied to the beam are all vertical, the 
value of M will be the same at whatever point of the section 



GExNERAL PRINCIPLES. 



97 



the origin is taken ; since the arm of each force will be the same 
for all origins in the same vertical line. If the loads and reactions 
are not all vertical, the value of M will generally depend upon 
what point in the section is taken as the origin of moments. 

117. Curves of Shear and Bending Moment. — The airve of 
shear for a beam is a curve whose abscissas are parallel to the 
axis of the beam, and whose ordinate at any point represents 
the external shear at the corresponding section of the beam. 

Let AB (Fig. 44) represent a beam loaded in any manner, and 
let A^B' be taken parallel to AB. At every point of AB^ 
suppose an ordinate drawn whose length shall represent the 
external shear at the corresponding 
section of AB. The line ab, join- 
ing the extremities of all these ordi- 
nates, is the shear curve. Positive 
values of the shear may be repre- 
sented by ordinates drawn upward 
from AB\ and negative values by 
ordinates drawn downward. (Instead of drawing AB^ parallel 
to the beam, it may be any other straight line whose extremi- 
ties are in vertical lines through A and B) 

The curve of bending mome^it for a beam is a curve whose 
abscissas are parallel to the axis of the beam, and whose ordinate 
at any point represents the bending moment at the correspond- 
ing section of the beam. Thus in Fig. 44, A"C'B" may repre- 
sent the bending moment curve for the beam AB. (Evidently 
A'B" might be inclined to the direction of AB, without destroy- 
ing the meaning of the curve.) Positive and negative values of 
the bending moment will be distinguished by drawing the 
ordinates representing the former upward and those represent- 
ing the latter downward from the line of reference A"B". 

1 18. Moment Curve a Funicular Polygon. — If the loads and 
reactions upon the beam are all vertical, every funicular poly- 
gon for these forces taken consecutively along the beam, is a 




98 



GRAPHIC STATICS. 



curve of bending moments. Thus, let MN (Fig. 45) represent 
a beam under vertical loads, supported at the ends by vertical 
reactions. Draw a funicular polygon for the loads and reactions 
as shown. 




-_;:;^5>o 



Fis. 45 



Now, by definition, the bending moment at any section is 
equal to the moment of the resultant of all external forces act- 
ing on the beam to the left of the section, the origin of 
moments being taken in the section. By Art. 56, this moment 
can be found by drawing through the section a vertical line and 
finding the distance intercepted on it by the two strings corre- 
sponding to the resultant mentioned ; the product of this inter- 
cept by the pole distance is the required moment. Hence, if 
oe (Fig. 45) is taken as axis of abscissas, the broken line made 
up of oa, oby oc, od, is a *' curve of bending moments." 

119. Design of Beams. — The principles involved in the 
design of beams will not be here fully discussed. Every prob- 
lem in design involves the determination of shears and bend- 
ing moments throughout the beam ; and the graphic methods 
of determining these will alone be considered in the following 
articles. 



§ 2. Beam Sustaining Fixed Loads. 

120. Shear Curve for Beam Supported at Ends. — Let MN 

(Fig. 45) represent a beam supported at the ends and sustain- 
ing loads as shown. Draw the force polygon ABCD, and with 
pole O draw the funicular polygon. The closing line is oe 



BEAM SUSTAINING FIXED LOADS. 99 

(marked also M^'N'^), and OE drawn parallel to J/'^V" fixes E, 
thus determining DE, EA, the reactions at the supports. 

Take M'N' as the axis of abscissas for the shear curve. 

The shear at every section can be at once taken from the 
force polygon. For, remembering the definition of external 
shear (Art. 1 1 5) we have : 

Shear at any section between J/ and P is EA (positive). 

Shear at any section between P and Q is EB (positive). 

Shear at any section between Q and R is EC (negative). 

Shear at any section between R and N is ED (negative). 

Hence the shear curve is the broken line drawn in the 
figure. 

121. Moment Curve for Beam Supported at Ends. — As in 
Art. 118, it is seen that the funicular polygon already drawn 
(Fig. 45) is a bending moment curve for the given forces. For 
the bending moment at any section of the beam is equal to the 
corresponding ordinate of this polygon, multiplied by the pole 
distance. 

For simplicity, it will be well always to choose the pole so 
that the pole distance represents some simple number of force- 
units. 

The sign of the bending moment is readily seen to be nega- 
tive everywhere, according to the convention already adopted 
(Art. 47). 

122. Shear and Moment Curves for Overhanging Beam. — 

Consider a beam such as shown in Fig. 46, supported at Q and 
T, and sustaining loads at M, P, R, S, and JV. 

This case may be treated just as the preceding, care being 
exercised to take all the external forces (loads and reactions) in 
order around the beam. 

The construction is shown in Fig. 46. First, the reactions at 
Q and T are found as in the preceding case, by drawing the 
funicular polygon, finding the closing line, and drawing OG 
parallel to it. The reactions are EG and GA. Then the value 



lOO 



GRAPHIC STATICS. 



of the external shear can be found for any section from the 
definition, and is always given in magnitude and sign by a cer- 
tain portion of the force polygon ABCDEFGA. The resulting 
curve is shown in the figure, M'N' being the line of reference. 



4^ ^1^ 



d d\e 



M 



M' 



t 



f 



a\g 



js Tf~> 



9\f 



M' 



nmri 



iiiii i iii i ii r 




N' 



N' 



:::::^^o 



Fig, 46 

Similarly, from the definition of bending moment, and the 
principle of Art. 56, the bending moment at any section is 
equal to the ordinate of the funicular polygon multiplied by the 
pole distance. The polygon is shaded in the figure to indicate 
the ordinates in question. It is seen that the bending moment 
is positive at every section of the beam. 

123. Distributed Loads. — In all the cases thus far discussed, 
the loads applied to the beam have been considered as concen- 
trated at a finite number of points. That is, it has been 
assumed that a finite load is applied to the beam ^t a point. 
Such a condition cannot strictly be realized, every load being 
in fact distributed along a small length of the beam. If the 
length along which a load is distributed is very small, no im- 
portant error results from considering it as applied at a point. 

In certain cases a beam may have to sustain a load which is 
distributed over a considerable part of its length, or over the 
whole length. Such a load may be treated graphically with 
sufficient correctness by dividing the length into parts, and 



BEAM SUSTAINING iMOVING LOADS. jqi 

assuming the whole load on any part to be concentrated at a 
point. The smaller these parts, the more nearly correct will 
the results be. 

It may be remarked by way of comparison that while algebraic 
methods are most readily applicable to the case of distributed 
loads, the reverse is true of graphic methods, which are most 
easily applied in the very cases in which analytic methods 
become most perplexing. 

124. Design of Beam with Fixed Loads. — The above exam- 
ples are sufficient to explain the method of treating any simple 
beam under fixed loads. Under such loading the shear and 
bending moment at each section of the beam remain unchanged 
in value, and no further discussion is necessary as a preliminary 
to the design of the beam. We proceed next to the case of 
beams with moving loads. 

§ 3. Beam Siistaming Moving Loads. 

125. Curves of Maximum Shear and Moment. — When a beam 
sustains moving loads, the shear and moment at any section do 
not remain constant for all positions of the loads. In such a 
case it is the greatest shear or moment in each section that is 
to be used in designing the beam. 

A curve of maximtcin shear is a curve of which the ordinate 
at each point represents the greatest possible value of the shear 
at the corresponding section of the beam for any position of the 
loads. 

A curve of maximum moment is a curve whose ordinate at 
each point represents the greatest possible value of the bending 
moment at the corresponding section of the beam for any posi- 
tion of the loads. 

In the following articles will be explained a method of deter- 
mining any number of points of the curves just defined, in the 
case of a simple beam supported at the ends. The moving 



I02 GRAPHIC STATICS. 

load will be taken to consist of a series of concentrated loads 
with lines of actions at fixed distances apart. An example of 
such a load is the weight of a locomotive and train ; the 
points of application of the loads being always under the several 
wheels. 

126. Position of Loads for Greatest Shear at Any Section. — 

In Fig. 47 (PI. Ill) let the vertical lines ab^ be, etc., represent 
the lines of action of the moving loads in their true relative 
positions ; the magnitudes of the loads being shown in the force 
polygon or "load line" ABCD . . . Let XY represent the 
length of the beam, and let it be required to determine the 
position of the moving loads that will cause the maximum posi- 
tive shear at any section, as at that marked /. 

First, consider the effect of a single load in any position. It 
is seen that any load to the right of y produces at that point a 
positive shear. For, by definition, the external shear is equal 
to the algebraic sum of all vertical forces to the left of the 
section, upward forces being reckoned positive. Now a load 
to the right of ^ causes no forces to act at the left of this 
section, except an upward reaction at X. On the other hand, 
a load to the left of y produces negative shear at that point. 
For the shear due to it is equal to the reaction at X minus the 
load itself, which will always be negative, since the load is 
greater than the reaction due to it. (It is also to be noticed 
that the shear due to a load at the left of the section is equal 
numerically to the reaction it produces at V.) 

Again, the shear due to any load is greater in magnitude, the 
nearer the load is to the section considered. For, as a load on 
J^V approaches y, the reaction at X increases ; and as a load 
on J(/ approaches y, the reaction at F increases. 

Two principles are thus reached to guide in assigning the 
position of the loads that will produce the greatest positive 
shear at any section : (i) The segment of the beam to the left 
of the section should be without load and that to the right fully 



BEAM SUSTAINING MOVING LOADS. 103 

loaded ; and (2) the loads on the loaded segment should be as 
near the section as possible. 

It would seem, therefore, that the greatest positive shear at 
the point y will occur when the loads are brought on the beam 
from the right to such a position that the foremost load in the 
series is just at the right of J. And, in general, this will be 
true, unless the foremost load is small in comparison with the 
whole load on the beam, or unless the distance between the first 
and second loads is considerable. In such cases it may be that 
a greater shear will result when the second load is brought up 
to the section. For, suppose the first load to be just at the 
right of J, and consider the effect of moving the whole series 
of loads to the left. As the first load passes the section, it 
produces a diminution of the shear equal in amount to the load. 
On the other hand, as the loads continue to move to the left, 
the effect of every load in producing reaction at X increases, 
while no further decrease in the shear at J occurs until the 
second load passes the section. Now, it may be that the net 
result of bringing the second load up to J will be to increase 
the shear at that section. If the first two or more loads are 
very small, it may be that the third or fourth load should be 
brought to the section to produce the greatest shear. Usually, 
however, it will be necessary to try only two (or at most three) 
positions. 

Instead of resorting to trials to determine for which position 
the shear is greatest, we may apply a simple rule which will 
now be deduced. 

Let /*i = magnitude of foremost load, /'.j^niagnitude of sec- 
ond load, etc., W being the total load on the beam. Let /= 
total span, and 7;^ = distance between Px and P.. Let -t' = dis- 
tance from Y to the center of gravity of W when P^ is at a 
given section. Let us compute the shear when P^ is just at 
the right of the section, and then determine the effect of moving 
all loads to the left, until the second load comes to the section. 

When P is at the right of the section, the shear is equal to 



I04 GRAPHIC STATICS. 

the reaction at X, say R. Then (calHng V the external shear) 
we have 

If now the loads be moved until P.y_ comes to a point just 
at the right of the section, the reaction due to W becomes 

W- -, and the shear at the section becomes 

The increase of the shear is, therefore, 

W{x-\-m) r> Wx Win r, 

T ^'~^-^~^'- 

This increase is plus or minus according as is greater or 

P I 
less than P^ ; that is, as ^Fis greater or less than — ^. Hence 

ni 

the following rule : 

The maxinittin positive shear in any section of the beam occurs 
when the foremost load is infinitely near the section, provided W 

PI PI 

is not greater than ^-. If W is greater tJian —^, the greatest 

m m 

shear will occur zvhen some succeeding load is at tJie section. 

In the above discussion it has been assumed that in bringing 
Pi up to the section no additional loads are brought upon the 
beam. If this assumption is not true, let W^ be the new load 
brought on the beam, and x^ the distance of its center of 
gravity from the right support when P^ is at the section. Then 
the shear corresponding to this position is 

and. the increase of shear due to the assumed change in the 
position of the load is 

Wm , W^x' o 



BEAM SUSTAINING MOVING LOADS. 



105 



Hence, in the statement of the above rule, we have only to 

substitute Wm+W'x' ior IVm; or, W^^-^ for W. The 

in 

additional load W^ may be neglected except when the condition 

P I 

W = — - is nearly satisfied; for the term Wx' will always be 

small in comparison with IVm. 

The application of this rule is very easy, and will save much 
labor in the graphic construction of the shear curve. 

If it is found that the first load should be past the section 
for the position of greatest shear, we may determine whether 
the second or the third should be at the section by an exactly 
similar method. We have only to apply the above rule, substi- 
tuting P2 for Pi, and for w the distance between P and P^. 

127. Determination of Maximum Shear. — The shear at any 
section due to any position of the moving loads can readily be 
determined from the force and funicular polygons, by a method 
which will now be shown. 

Draw in succession the lines of action of the loads at their 
proper distances apart, as in Fig. 47 (PI. III). Draw the load- 
line ABCD . . ., and choosing a pole O^ draw the funicular 
polygon. The same space diagram may be used for any position 
of the moving loads ; for, instead of moving the loads in either 
direction, we may assume the loads fixed and move the beam 
in the opposite direction. 

By applying the rule deduced in Art. 126, it may be shown 
that there is a point Z in the beam, such that for any section 
to the right of Z the foremost load should be at the section in 
order that the shear may be a maximum ; while for any section 
to the left of Z the second load must be brought to the section. 
The position of this point is readily determined, as follows : 

In the case shown in Fig. 47, we have 

/=64ft., m = S.i ft., 7^1 = 8000 lbs., ^=-^ x 8000 = 63200. 

m 8.1 

Hence, as the load moves from right to left, W is less than 



I06 GRAPHIC STATICS. 

PI. 

— — until the whole load on the beam reaches 63200 lbs. This 

VI 

occurs when the fifth load is at Y. Hence the point Z is 
located at a distance from F equal to the distance between the 
lines of action of the first and fifth loads. 

For any point to the left of Z, the greatest shear will prob- 
ably occur when the second load is at the section. To test 
whether in any case the third load should be at the section, we 
apply the same principle, as follows : The second load is 1 5000 
lbs. ; the distance between the second and third loads is 5.8 ft. 
Hence, for greatest shear, the third load should not be at 
the section unless the whole load on the beam is at least 

-^x 15000 lbs. or 165500 lbs. But it is easily seen that the 

total load can never be so great. 

We may now explain the construction for finding the great- 
est shear at any section. Consider a section at the left of Z, 
as at the point marked y! From the above reasoning it is evi- 
dent that the second load must be brought close to the section. 
In Fig. 47, SjSj is drawn to represent the beam, its position 
being such that the point J is infinitely near bc^ the line of 
action of the second load. Through the extremities of the 
beam S^Sj draw vertical lines to represent the action-lines of 
the reactions at the supports. The portion AB^ of the load 
line represents the loads now on the beam, and the correspond- 
ing extreme strings of the funicular polygon are oa and ob\ If 
now we consider the funicular polygon for the loads and reac- 
tions acting on the beam in the supposed position, we see that 
the closing side is found by joining the points in which the 
action-lines of the end reactions intersect oa and ob^ respec- 
tively. This is marked of in the figure. Now draw in the 
force diagram a ray parallel to this closing line, and lety be its 
point of intersection with the load line; then B^J^ smd/'A are 
the two reactions. The shear at the point _/ is evidently the 
algebraic sum of the reaction /'A and the load AB ; hence it 



BEAM SUSTAINING MOVING LOADS. 



107 



is represented hyJ'B. In Fig. 47 let X, K- be drawn to repre- 
sent the length of the beam, and let the ordinates of the shear 
curve be drawn from it. Then from y we lay off an ordinate 
equal toJ^B. The whole curve of maximum shear is shown in 
the figure, but the construction for finding it is given for only 
one point. 

128. Shear Curve for Combined Fixed and Moving Loads. — 

Let the beam sustain a fixed load of 25000 lbs. uniformly dis- 
tributed along the beam. The shear close to the left support 
due to this load is equal to the reaction, or 12500 lbs. ; and 
decreases as we pass to the right by ^^-^-- lbs. for each foot. 
At the middle of the beam the shear is zero, and at the right 
support it is —12500 lbs. Hence the shear curve is a straight 
line, and may be drawn as follows : From X^ lay off an ordinate 
downward representing 12500 lbs., and from Ys an ordinate 
upward representing 12500 lbs.; the straight line joining the 
extremities of these ordinates is the shear curve for the fixed 
load. Positive shears are laid off downward and negative 
shears upward for the reason that, if this be done, the greatest 
positive shear at any point due to fixed and moving loads is 
represented by the total ordinate measured between the shear 
curves for fixed loads and for moving loads. It is seen that at 
a certain point somewhere to the right of the center of the 
beam this greatest shear is zero, and for all sections to the 
right of this point it is negative. This point is determined by 
the intersection of the two shear curves. 

1 29. Position of Loads for Greatest Bending Moment. — 

Referring to the beam A^F shown in Fig. 47 (PL III), consider 
the bending moment at any section (as^) due to a load any- 
where on the beam. First, suppose the load is at the right of 
the section. In this case the only force brought upon the por- 
tion of the beam to the left of the section is a reaction at X. 
By definition (Art. 116) the bending moment atyis the moment 
of this reaction about an origin aty, and is negative. Moreover, 



^ 



I08 GRAPHIC STATICS. 

since the reaction will become greater as the load moves toward 
the left, the greatest bending moment due to a load at the right 
of the section will occur when the load is as near the section as 
possible. Next consider a load at the left of ihe section. The 
bending moment due to it is equal to the moment of the result- 
ant of the load and the left reaction. But this resultant is 
equal and opposite to the reaction at the right support, and has 
the same line of action (because the load and the two reactions 
due to it are in equilibrium) ; hence the bending moment is the 
negative of the moment of the right reaction about an origin 
in the section. The bending moment is therefore negative, 
and is greatest when the right reaction is greatest. Hence, a 
load at the left of the section produces its greatest bending 
moment when the load is as near as possible to the section. 

We are therefore led to the following general principles for a 
simple beam supported at the ends : 

(i) The bending moment at every section is negative for all 
positions of the moving loads. 

(2) The negative bending moment at any section has its 
greatest value when the whole beam is loaded as completely as 
possible, and the heaviest loads are near the section. 

These principles serve as a general guide, but unless the 
moving loads are equidistant and equal in magnitude it will be 
necessary to try several positions before the position for great- 
est bending moment will be known. Usually one of the 
heaviest loads should be directly at the section. 

Instead of resorting to repeated trials, the position of loads 
giving greatest bending moment at any section may be found 
by means of a simple rule which will now be deduced. 

Let /= length of beam between supports; JV= total load on 
beam; ;tr— distance of center of gravity of W from right sup- 
port; J'Fi^: load on beam to left of given section; ,r = distance 
of center of gravity of Wi from section; /i=:distance of sec- 
tion from left support ; R = reaction at left support ; M= 
bending moment at the section. Then we have 



BEAM SUSTAINING MOVING LOADS. 109 



IV/ 
M= Rk - W,x\ = ^x- IVrX'i, 

If the loads all move an infinitesimal distance to the left, 
then X and Xi receive equal infinitesimal increments, and the 
increment of Mis 

dM= ^dx- IV.dx, = (^ - W\ dx 

(since dx^ = dx). Now if M is a maximum, we must have 

= 0; that IS, i— lVi = o; or, -— ^ = — -. Hence the follow- 

dx I l\ I 

ing rule : 

Whe7i the bending moment at any section of the beam has its 

greatest value, the loads oji the two segments of the bea^n are to 

each other as the lengths of the segments. 

In case of concentrated loads, the condition just stated can 
generally not be exactly satisfied except for certain sections of 
the beam. It will usually be found that the position most 
nearly fulfilling the condition is that in which some heavy load 
is just at the section. This will be illustrated in the next 
article. 

[Note. — The above reasoning applies only to the case of concentrated loads, and 
is not rigorous for this case, if, in the position of maximum moment for any section, 

a load is at the section ; for J will not then be zero, as assumed above. In such 

a case -^— is discontinuous, and its value changes sign as the loads pass through the 

TAT" 1 /f/^ 

position for which — ^ =0. It is still true that this condition is satisfied in 

the position of maximum moment, provided the load at the section is regarded as 
divided in some certain ratio between the two segments of the beam.] 

130. Determination of Bending Moments. — The application 
of the above principles and the method of determining the 
bending moment at any section of the beam will now be shown 
for the case represented in Fig. 47 (PI. III). Let the greatest 
bending moment be found for the section J. From the general 



no GRAPHIC STATICS. 

principles deduced in Art. 129, it is probable that the loads 
nearest the section should be the third and fourth in the series. 
Also, since the point _/ divides the span into segments of 16 ft. 
and 48 ft., the load on the left segment should be one-fourth 
the total load on the beam. Now it is readily found that when 
the load CD is just at the right of the section, the whole load 
on the beam is 112,000 lbs., while the load on the left segment 
is 23,000 lbs., which is less than one-fourth of 112,000. And 
when the load CD is just at the left of the section, the load on 
the left segment is 38,000 lbs., which is greater than one-fourth 
of 112,000. Hence the bending moment is a maximum when 
the load CD is just at the section. 

In Fig. 47, MjMj represents the beam for this position of the 
loads, and the closing line 01 the funicular polygon is marked 
oj" . Using the principle of Art. 56, we find the bending mo- 
ment at the given section by measuring the distance intercepted 
on ^^by the strings od and oj" and multiplying it by the pole 
distance in the force diagram. 

Let Xra Vm (Fig- 4/) be the axis of abscissas for the curve of 
maximum moments. From the point y draw an ordinate equal 
to the intercept just found ; this locates a point of the required 
moment curve. 

Other points may be determined in the same manner. The 
curve is shown in the figure, but the construction is not given 
for any section except that at /. 

It must be remembered that each ordinate is to be multiplied 
by the pole distance. Hence it will be convenient to choose 
the pole distance equal to some round number of force units. 

131. Moment Curve for Fixed Loads. — The greatest bending 
moment due to moving loads must be combined with the bend- 
ing moment due to fixed loads. If the fixed load is uniformly 
distributed, as already assumed in the computation of shear, 
it may be divided into parts, each assumed concentrated at 
its center of gravity, and a funicular polygon drawn, using the 



BEAM SUSTAINING MOVING LOADS. m 

same pole distance employed in the force diagram for moving 
loads. The ordinates of this funicular polygon may be laid off 
upward from the line X^Vm, and their ends joined to form a 
continuous curve. The total ordinate from this curve to that 
already drawn for moving loads represents the true greatest 
bending moment at the corresponding section of the beam. 
The curve is shown in the figure, but the construction is 
omitted, since it involves no new principle. 

It is to be remembered that the bending moment found for 
any section is a possible value for the other section equally 
distant from the center of the beam, since the train may be 
headed in the opposite direction, and the same construction 
made, viewing the beam from the other side. The same state- 
ment holds as to shears. 

132. Design of Beam sustaining Moving Loads. — In designing 
a beam to sustain moving loads, the greatest shear and bending 
moment that, can come upon it for any position of the loads must 
be known for every section. The methods above given are 
sufficient for the determination of these quantities ; and the 
problem of designing the beam will not be here further discussed. 

133. Plate Girders and Lattice Girders. — A p/ate girder is a 
beam built up of rolled plates and angle-irons riveted together, 
the cross-section being as shown in Fig. 48. If 
latticing is substituted for the plate, the beam be- ^ 
comes a lattice girder. Railway bridges of spans 
under 100 ft. frequently employ either rolled beams, 
plate girders, or lattice girders. (See Cooper's 
"■ Specifications for Iron and Steel Railroad 
Bridges.") The methods given in the preceding 
articles are especially useful in designing this class 
of bridges. cM=i 

The student should make the complete construe- fis-^s 
tion for determining the curves of maximum shear and bending 
moment for a beam designed to carry the series of moving loads 
shown in Fig. 47. 



CHAPTER VII. TRUSSES SUSTAINING MOVING 

LOADS. 

§ I. Bridge Loads. 

1 34. General Statement. — The most important class of trusses 
sustaining moving loads is that of bridge trusses. The two 
main classes of bridges are highway bridges and railway bridges. 
The forms of trusses most commonly used differ for the two 
classes, as do also the amount and distribution of loads. 

Before the design can be correctly made, the weights of the 
trusses themselves must be known. Since these weights de- 
pend upon the dimensions of the truss members, they cannot 
be known with certainty until the design is completed. The 
remarks made in Art. 82 regarding the design of roof trusses 
are here applicable. 

In the following articles we shall give data available for 
preliminary estimates of truss weights. 

As stated in Art. 133, railway bridges of short span are fre- 
quently supported by rolled or built beams. When the span is 
longer than 100 ft., trusses should be used. (Cooper's "Speci- 
fications.") 

135. Loads on Highway Bridges. — (i) Permanent load. — 
The permanent load sustained by a highway bridge truss in- 
cludes the weight of the truss itself, of the lateral or '^ sway " 
bracing, of the floor and the beams and stringers supporting it. 
These weights are all subject to much variation, but, for pur- 
poses of preliminary design, the following formula, taken from 
Merriman's ''Roofs and Bridges," may be used. 

112 



BRIDGE LOADS. 



113 



Let «; = weight of bridge in pounds per linear foot ; /=:span 
in feet ; (^ = width in feet. Then 

w=\^o-^\2 b^o.2 bl—o.^ I. 

(2) Snow load. — The weight of snow may be taken as in 
case of roof trusses (Art. 84). The values there given are 
probably in excess of those ordinarily employed in practice. 

(3) Wind load. — The pressure of wind striking the bridge 
laterally is resisted by the chord members together with the 
lateral bracing. These constitute horizontal trusses, in which 
the stresses are to be found in the same way as for the main 
trusses of the bridge. As the determination of wind stresses 
requires the use of no special methods or principles, they will 
not be here considered. The student is referred to Burr's 
" Stresses in Roofs and Bridges," Merriman's " Roofs and 
Bridges," and other available works for a complete discussion 
of wind pressure and its effects on bridge trusses. 

(4) Moving load. — The most dangerous moving load for a 
highway bridge is usually a crowd of people or a drove of 
animals. This is commonly taken as a uniformly distributed 
load, which may cover the whole bridge or any portion of it. 
Its value is variously taken at from 60 lbs. to 100 lbs. per square 
foot of area of floor, depending upon the span and upon local 
conditions. 

It may be that in certain cases the greatest stresses will 
result from the passage of heavy pieces of machinery over the 
bridge, as, for example, a steam road roller. This should of 
course be considered in the design. 

For a complete discussion of loads on highway bridges, the 
student is referred to Waddell'c '' Highway Bridges." 

136. Loads on Railway Bridges. — (i) Permanent load. — 
The permanent load on a railway bridge includes {a) the weight 
of the track system, which is known or may be determined at 
the outset ; {b) the weight of longitudinal stringers and cross- 



114 GRAPHIC STATICS. 

beams, which can be determined before the trusses are designed ; 
and {c) the weights of trusses. The weight of the track system 
may be taken at 400 lbs. per linear foot for a single track. 
(See Burr's " Stresses in Bridge and Roof Trusses " ; Cooper's 
'' Specifications for Iron and Steel Railroad Bridges " ; Merri- 
man's '' Roofs and Bridges.") The total weight of track system 
and supporting beams and stringers varies from 450 lbs. to 
600 lbs. per linear foot. (Merriman.) For spans less than 100 
feet, Merriman gives the following formulas, in which zv is the 
total dead load of the bridge in pounds per linear foot, and / is 
the span in feet : 

For single track, w= 560+ 5.6 /. 

For double track, w= 1070-r 10.7 /. 

See also Art. 8 of Burr's work above cited. ^ 

(2) Snow and zvind. — Railway bridges usually offer little 
opportunity for the accumulation of snow. Wind pressure is, 
however, an important factor. Besides the pressure upon the 
bridge itself, the pressure upon trains crossing the bridge must 
be considered. The latter is a moving load and may be dealt 
with in the same way as other moving loads. Its amount may 
be computed from the area of the exposed surface of the train 
and the known (or assumed) greatest pressure due to wind 
striking a vertical surface (Art. 85). 

For further discussion of wind pressure, the student is referred 
to the works already cited. The graphic methods of deter- 
mining stresses due to wind will be evident when the methods 
for vertical loads given in the following articles are understood. 

(3) Moving loads. — The moving load to be supported by a 
bridge consists of the weights of trains. Such a load is applied 
to the track at a series of points, namely, the points of contact 
of the wheels. But the load is applied to the trusses only at 
the points at which the floor beams are supported. Hence the 
actual distribution of loads upon the truss is somewhat com- 
plex. It was formerly common to substitute for the actual 



TRUSS REGARDED AS A BEAM. 



115 



load a uniformly distributed load, thus simplifying the problem 
of determining stresses. It is now more usual to consider the 
actual distribution of loads for some standard type of locomo- 
tive used by the railroad concerned, or specified by its engi- 
neers. For examples of such distributions the student is 
referred to Cooper's " Specifications " already cited ; also to 
Fig. 47, and to the following portions of this chapter. 

137. Through and Deck Bridges. — A bridge is called throiigJi 
or deck according as the floor system is supported at points of 
the lower or of the upper chord. In the former case, if the 
trusses are too low to require lateral bracing above, they are 
called /d?wj/ trusses. 

The weight of the truss itself is to be divided between the 
upper and lower joints. But the weight of the floor system 
and of the supporting beams and stringers comes wholly at the 
lower joints of a through bridge, or at the upper joints of a 
deck bridge. The moving load is, of course, applied at the 
same joints at which the floor system is supported. 

If the floor system is supported directly upon the upper 
chord, as is sometimes the case, the moving load and part of 
the dead load produce bending in the chord members ; other- 
wise the design is unaffected by this construction. 

§ 2. Truss Regarded as a Beam. 

138. Classification of Trusses. — Since a bridge truss acts as 
a practically rigid body resting on supports at the ends or other 
points and sustaining vertical loads, it may be regarded as a 
beam, and trusses may be classified in the same way as beams 
(Art. 114). The only class to be here considered is that of 
simple trusses, or such as may be regarded as rigid bodies for 
the purpose of determining the reactions. 

Cantilever trusses and continuous trusses are defined like the 
corresponding classes of beams (Art. 114). The most impor- 
tant case is that of a truss simply supported at the ends. 



ii6 



GRAPHIC STATICS'. 



139. External Shear for a Truss. — If a truss be regarded as 
a beam, the external shear, resisting shear, and internal shear- 
ing stress at any section may be defined just as in Art. 115. 
In some forms of truss a knowledge of the external shear at 
any section makes it possible to compute readily the stresses in 
certain truss members. Thus, in the portion of a truss repre- 
sented in Fig. 49, let the section 
MN cut three members, of which 
two are horizontal. (The member 
x^y is disregarded.) Since each 
member can exert forces only in 
the direction of its length, the ex- 
ternal shear in the section MN 
must be wholly resisted by the di- 
agonal xy ; and the internal force 
in xy must be such that its resolved 
part in the vertical direction is equal in magnitude to the exter- 
nal shear in the section. Let the external shear V be repre- 
sented by YZ (Fig. 49) ; draw YX parallel to yx and ZX 
horizontal; then XFwill represent the stress in xy. If F is 
positive (Art. 115), the stress in xy is a tension. If J^is nega- 
tive, the stress is a compression. If the member xy were 
replaced by one sloping the other way from the vertical, these 
statements as to kind of stress would be reversed. 

If no two of the three members cut by any section are par- 
allel, the stresses cannot be computed so simply, since all may 
contribute components of force to resist the external shear. 




Fig. 49 



140. Bending Moment for a Truss. — In many cases the 
stresses in the truss members can be found from the values of 
the bending moment at different sections. 

Thus, in Fig. 49, let a section MN be taken cutting three 
members as shown, and let the origin of moments be taken at 
the point of intersection of two of them (as bx and xy) ; then 
since the moments of the internal forces in these two will be 



PARALLEL CHORDS — CONCENTRATED LOADS. 



117 



zero, the resisting moment is equal to the moment of the inter- 
nal force in the third member ay. The arm of this latter force 
is the perpendicular distance of its action line from the origin. 
Call it k, and let/ represent the force itself, and i^^the bending 
moment. Then, numerically, 

h 

If M is positive (in accordance with the convention of Art. 
47), / must act from right to left, hence the stress in ay is a 
compression. If M is negative, the stress is a tension. (It 
must be remembered that the forces whose moments make up 
the bending moment M act upon the portion of the truss to 
the left of the section.) 

This method applies equally well to the case in which no two 
of the members cut are parallel. It is an application of a prin- 
ciple of much importance in the determination of stresses in 
truss members. 



§ 3. Truss zvitJi Parallel CJiords siistaming Concentrated 

Loads. 

141. General Method. — It was shown in Art. 139 that when 
both chords of the truss are horizontal, so that a vertical sec- 
tion through the truss will cut but one inclined member, the 
stress in such an inclined member may be easily found from the 
external shear in the section. Hence for such a truss, we may 
determine the greatest stresses in the web members by con- 
structing the curve of maximum shear for the truss, in a man- 
ner similar to that employed in Art. 127 for a beam. 

Again, the curve of maximum bending moments will usually 
enable us to determine the greatest chord-stresses (Art. 140). 

The construction of these curves for the case of a truss is 
not quite so simple as in case of a beam, for reasons to be 
explained in the following article. 



Il8 GRAPHIC STATICS. 

142. Distribution of Loads on Truss. — If the curves of maxi- 
mum shear and moment for a given series of concentrated loads 
be constructed as in Arts. 127 and 130, they will not represent 
correctly the maximum shear and moment at all points of the 
truss. The reason for this will be seen by considering the way 
in which the loads are actually applied to the truss. The road- 
way of the bridge is usually supported by longitudinal beams, 
which themselves rest upon cross-beams or girders, the latter 
being supported at the joints of the truss. Hence, however 
the loads may be distributed upon the roadway, they are borne 
by the truss only at the joints. But in the method used in 
Arts. 127 and 130 for drawing curves of shear and bending 
moment, it was assumed that every load rests directly upon the 
beam, at a point immediately under the load. 

In the case of the truss, since loads are applied only at the 
joints, the shear at any instant is the same throughout the 
length of a panel. What is needed, therefore, is the greatest 
possible shear for each panel of the truss. For simphcity, we 
shall consider first the shear due to dead loads, and then that 
due to the live loads, taking a numerical example. 

143. Problem — Numerical Data. — Assume a ''through" 
truss of the form shown in skeleton in Fig. 50 (PI. IV) ; let the 
span be 96 ft., the truss being divided into 6 panels of 16 ft. 
each. The depth is taken at 15 ft. Assume the dead load as 
1 100 lbs. per linear foot for the whole bridge, — one-half being 
borne by each truss, — of which one-third is the weight of the 
floor system and track, and is therefore borne by the lower 
chord, and the remaining two-thirds is assumed to be divided 
equally between the upper and lower chords. We have then 
for the dead loads : 

Lower panel load = 5 866 lbs., say 6000 lbs. 
Upper panel load = 2933 lbs., say 3000 lbs. 

For the moving load we will " assume a train drawn by two 
locomotives with weights distributed as shown in Fig. 50. In 



PARALLEL CHORDS —CONCENTRATED LOADS. 



119 



computing stresses the train will be supposed to come on the 
bridge from the right. 

144. Shear Curve for Fixed Loads. — The curve of shears for 
the dead load may be constructed as follows : In Fig. 50 (PL IV), 
the line X,V, represents the axis of abscissas for the shear 
curve. The reaction at the left support, minus the dead load 
borne at that point, gives 24000 lbs. as the positive shear for 
every section between the support and the first upper panel 
joint. This is laid off downward from X,V,. Then passing to 
the right, as each load is passed, its amount must be sub- 
tracted from the shear. The result is the broken line X' V 
shown in the figure. 

145. Moment Curve for Fixed Loads. — To construct the 
bending moment curve for dead loads, we have only to draw a 
funicular polygon for these loads and the reactions due to them ; 
the vertical ordinate at any section will represent the bending 
moment at that section. The actual value of the bending 
moment is the product of the ordinate into the pole distance ; 
the former being measured in linear units, the latter in force 
units. The curve is shown in Fig. 50 (PI. IV), the ordinates 
being laid off upward from the line X^ Vm- The force polygon 
is shown at IV, the pole being O, taken opposite the middle 
point of the load-line in order that the closing side of the funic- 
ular polygon may be horizontal, and can therefore be made to 
coincide with X^Y^. The pole distance represents 120,000 lbs. 

146. Actual Shear for Given Position of Moving Loads. — 

The shear in any panel is readily found when the position of 
live loads is known. We have only to apply the definition of 
external shear (Art. 115) as the sum of all vertical forces acting 
on the portion of the truss to the left of the section con- 
sidered. That is, we must find the reaction at the left abut- 
ment and subtract from it the sum of the loads between the 
left abutment and the section. It must be noticed that any 



I20 GRAPHIC STATICS. 

load on the panel considered is divided between the two adja- 
cent joints; the part supported at the left end of the panel is 
to be subtracted in computing the shear. The graphic con- 
struction is explained in the following articles. 

[Note. — It will be noticed that the actual distribution of any load among the 
different panel points is not strictly determinate by methods heretofore treated, if the 
longitudinal beams supporting the floor system are continuous (supported at more 
than two points. See Art. 114). We shall assume the portion of such a beam 
between two supports to act as a simple beam. The results of this assumption are 
probably as reliable as could be obtained by a more elaborate discussion.] 

147. Position of Moving Load for Maximum Shear in Any 
Panel. — The position of the moving load which will give the 
maximum shear in any panel may be determined graphically by 
repeated trials. But it is possible to deduce a simple rule which 
will shorten the labor very materially. 

It must first be noticed that the general principles stated 
in Art. 126, in treating of beams, are equally applicable here. 
We know, therefore, in a general way, that in order to get the 
greatest positive shear in a given panel, the portion of the truss 
to the right should be loaded as completely as possible, while 
the part to the left should be free from loads. This rule is not, 
however, absolute. In general the foremost loads will be upon 
the panel in question, or possibly to the left of it, in the posi- 
tion of greatest shear. We shall now prove the following 

Proposition. — When the shear in any panel has its greatest 
value, the load upon that panel is to the total load on the truss 
as the length of the panel is to the length of the truss. 

Let XF(Fig. 51) represent the length of the truss ( = /), and 
CD ( = /') the length of the panel. Let W — total load on 
truss, and W^ — total load on the panel. Consider the effect of 
moving all the loads to the left. Evidently the reaction at X 
is increased and the load at C is increased. The former 
increases the shear in the panel, while the latter decreases it. 
These are the only changes which are caused in the shear by 
the movement of the load. Now if the position of loads is 



PARALLEL CHORDS — CONCENTRATED LOADS. 



121 



such that the shear is a maximum, the two effects mentioned 
must just neutrahze each other for a very small displacement of 
the loads. (Otherwise a small displacement in one direction or 
the other would increase the shear.) 




Let X — distance of center of gravity of W from Fand x^ = 
distance of center of gravity of W^ from D. Let R — reaction 
at X and P = load at C due to IV. Then 



'X 



X 



If the whole load moves any distance, x and x^ both increase 
by the same amount. For an infinitesimal displacement. 



dR = — dx\ dP 



— dx = — — ax. 
i' /' 



Now the increment of the shear is dR — dP or 



W W'\ r 

dx. 

I V ) 



Since this must be zero (as stated above), we have 



W W 



or 



/' 



/ 



W /' 



for the position of maximum shear. The proposition is there- 
fore proved. 

The condition may be written W=—W\ If the truss is 

... I 

divided into n equal panels, then — = n, and W=ii W. 

This principle is so simple that it is easier of application than 
the graphic method of repeated trials. 



122 GRAPHIC STATICS. 

[Note. — In the above discussion it is assumed that IV and JV both remain con- 
stant during the movement c/jc; that is, that no load enters or leaves the panel or the 
whole truss. Hence the reasoning is not rigorous for a case in which the condition 

IV = — W requires a load to be at Cor D, or at X or Y. The remark made in the 
note in Art. 129 applies here. The condition W = ~ W may always be applied pro- 
vided a load at any joint may be treated as if divided between the adjacent panels in 
any desired ratio.] 

148. Construction of Curve of Maximum Shear. — In Fig. 50 
(PI. IV) are shown the lines of action of the series of moving 
loads to be carried by the truss. The force polygon or load 
line is drawn for the whole series of loads, being the line 
ABC. . . . For the purpose of drawing the funicular polygon, 
the uniformly distributed load is divided into parts of 6000 lbs., 
each assumed to act at its center of gravity. With a pole O 
(the pole distance being taken so as to represent 120000 lbs.) 
the funicular polygon is drawn for the series of loads. Let us 
now consider the greatest shear in any panel of the truss, as in 
KL (Fig. 50). 

Applying the principle already deduced, it is easily shown 
that the greatest shear occurs when the load CD is at the point 
Z. For, suppose CD is just at the left of L ; the total load on 
the panel is then 38000 lbs., while the total load on the truss 
is 172000 lbs. Now the ratio of the whole span to the panel 

length I—, Art. 147) is 6; and since 38000 lbs. is greater than 

one-sixth of 172000 lbs., the load on the panel is too great to 
satisfy the required condition. Again, suppose CD is just at 
the right of L ; the total load on the panel is 23000 lbs., which 
is less than one-sixth of 172000; hence the load on the panel is 
too small to satisfy the condition for maximum shear. We con- 
clude, therefore, that the shear is a maximum when the load 
CD is just at the point L. In Fig. 50, S2S2 represents the 
beam with load CD at L. Drawing vertical lines through the 
ends of vS2^'2, the points in which these intersect the strings oa 
and of determine two points of the closing string {0/2) of the 
funicular polygon for the loads and reactions now acting on the 



PARALLEL CHORDS — CONCENTRATED LOADS. 



123 



truss. The ray OJ2 is drawn parallel to this closing line, inter- 
secting the load line in J^. This determines J2A as the left 
reaction. To get the shear in the panel, we must subtract from 
this reaction the portion of the two loads AB, BC, which comes 
upon the truss at K. This portion is found by treating KL as 
a simple beam and determining the reactions at K and L due to 
loads AB and BC. Draw vertical lines through the points K 
and L of the truss (in the position S^S^ ; the points in which 
these verticals intersect the strings oa and oc are two points 
of the closing side of the funicular polygon for the loads 
and reactions on the beam KL. Parallel to this closing side 
draw a ray OJ2. This line divides the load line AC into two 
segments representing the portions of AB and BC supported 
on the truss at K and L respectively. Hence the shear in 
the panel is the reaction JoA minus the load represented by 
the line AJ2. 

Taking X, K (Fig. 50) as the axis of abscissas for the curve 
of shears, the portion of the curve of maximum shear corre- 
sponding to the panel KL is a straight line parallel to X^ Y„ at a 
distance from it equal to J^Ji. 

The curve of greatest shear for the whole truss is shown in 
the figure, being the broken line X" I^,, but the construction is 
given only for the panel KL. For other panels the method 
will be exactly similar. 

149. Position of Load for Maximum Bending Moment. — 

In computing stresses in the chord members, the greatest 
bending moment is needed for each vertical section containing 
a chord joint. In considering the position of moving loads 
which will give the greatest bending moment at any section, 
the sections through joints carrying live loads must be treated 
separately from those through joints free from loads. 

Let Fig. 51 represent a through truss (Art. 137), and con- 
sider first the bending moment at a section containing one 
of the lower joints, as C It is readily seen that for such a 



124 



GRAPHIC STATICS. 



section the bending moment due to a load anywhere on the 
truss is exactly the same as if the truss were a beam upon 
which the load was supported directly. For, consider a load on 
any panel as EC. Though the load is actually divided between 
B and C, the reaction it causes at X is the same as if the truss 
were a beam supporting the load directly. Also, since the load 
is the resultant of the two parts at B and C, the sum of the 
moments of these two parts about any origin is equal to 
the moment of the load itself. Now, in computing bending 
moment caused by the given load at any section, we take the 
moment of the left reaction minus the part of the load which 
is carried at the left of the section. If the section is anywhere 
except betzveen B and C, the result is the same whether we use 
the given load in its actual position, or its two components at 
B and C. 

For finding the greatest bending moments at the sections A, 
B, C, etc., we may therefore apply the same rule as in the case 
of a beam (Art. 129). That is, when the bending moment for 
such a section is a maximum, the loads on the two segments of 
the truss are proportioned to the lengths of the segments. 

Next consider a section somewhere between two loaded 
joints, as at C (Fig. 51). The principles above stated hold for 
such a section for loads anywhere except on the panel BC. In 
case of a load on this panel, only a part is borne at the left of 
C ; and, in computing bending moment, the part coming at C 
is to be omitted. Hence the above reasoning is not strictly 
applicable to this case. The following modification of the 
principle above deduced may however be shown to hold for any 
section of the truss. 

Let the given section divide a panel BC into segments BC^ 
and O^C. Then, if all loads on BC are treated as if divided 
betweefi B and C in the ratio of BC" to C"C, the general rule 
applicable to a beam may be used i7t assigning the position of 
loads for maximum bending moment. 

[The proof of this proposition will be omitted. The mode 



PARALLEL CHORDS — CONCENTRATED LOADS. 



125 



k 



of reasoning to be applied is similar to that used in Art. 147. 
The student should attempt the proof for himself.] 

In general, the error will be small if the bending moment 
curve for the case of a beam is used in computing bending 
moments at all panel joints of the truss. 

150. Curve of Maximum Bending Moments. — The greatest 
bending moment at any section can be easily found as soon 
as the position of moving loads producing it is known. The 
method will be explained with reference to the truss shown in 
Fig. 50 (PL IV). 

By Art. 149, the greatest bending moment for each of the 
points K, L, M, A'', P, is exactly the same as if the truss were 
a beam upon which the loads were supported directly. It is 
unnecessary to show the construction for these points, it being 
identical with that of Art. 130. We therefore consider a sec- 
tion through an upper joint as T. First the position of the 
moving load must be assigned. 

By Art. 149, we may apply the general principle that the 
loads in the two segments of the truss should be in the same 
ratio as their lengths, provided all loads on the panel MN are 
treated as if equally divided between 7]/ and A^ (since the sec- 
tion considered divides MN into equal parts). Again, since 
the segment of the truss to the left of the section is seven- 
twelfths of the whole span, the load on the left segment should 
be seven-twelfths the total load on the truss. 

It is found by trial that the condition for maximum bending 
moment at T is satisfied when the line of action b'c' is at M. 
For, the total load on the truss, when the load is near this posi- 
tion, is 182000 lbs. Now suppose the load B^O is just at the 
left of M. The three loads, C'D\ D'E', ET\ must be regarded 
as equally divided between J/ and N\ this gives for the whole 
load to the left of T 11 1500 lbs., which is greater than seven- 
twelfths of 182000 lbs. But if B'O is just at the right of M, 
half of it must be regarded as borne at N, and the v/hole load 



126 GRAPHIC STATICS. 

on the truss to the left of T is 104000 lbs., which is less than 
seven-twelfths of 182000 lbs. 

Now draw Mt Mt to represent the beam in the position for 
maximum bending moment at T; that is, with the load B'C at 
M. From its extremities draw vertical lines intersecting the 
strings od and oc", and through the points thus determined 
draw ot', the closing side of the funicular polygon for all the 
loads and reactions acting on the truss. 

In determining the bending moment by the method of Art. 
130, we must consider. the loads on the panel MN as replaced 
by their components at M and JV respectively. Draw vertical 
lines through J/ and 7\^ intersecting the strings oc' and of, and 
join the points thus determined. The line thus drawn should 
replace the corresponding portion of the funicular polygon 
before drawn {i.e., the strings oc', od', oe' , and a part of) ; 
because it is, in fact, a line of the funicular polygon which will 
be obtained if CD', D'E', and E'F' be replaced by their com- 
ponents which are actually applied to the truss at M and N. 
We now draw through T a vertical line intersecting ot' and the 
corrected string just determined. The intercept thus deter- 
mined, multiplied by the pole distance, gives the bending 
moment at T. 

From X„, V„t as a line of reference draw downward an ordi- 
nate at the point T equal to the intercept thus found ; this 
gives a point of the curve of maximum bending moments. 

Other points of the curve may be found in the same way. 
The whole curve is shown in the figure, but the construction is 
not given except for the section T. 

[Note. — The condition for maximum moment at a section may sometimes be 
satisfied for more than one position of the moving load. This is the case for the 
section at T; it will be found by trial that the condition is satisfied when the load 
C"D" is just at the right end of the truss. The bending moment for this position 
is, however, found to be slightly less than for the position above taken. In most 
cases, the position for greatest moment can be very nearly predicted by remembering 
that the truss should be loaded as completely as possible, and that the heavy loads 
should be as near the section as possible. These principles will also serve as a guide 
in distinguishing whether the moment is a maximum or a minimum when the condi- 
tion is satisfied.] 



PARALLEL CHORDS — CONCENTRATED LOADS. 



127 



It should be noticed that the value found for the bending 
moment at any section is a possible value for another section 
equally distant from the middle point of the truss, since the 
train may cross the bridge from the left instead of from the 
right. Hence for two points, as L and N, equally distant from 
the middle of the truss, the greater of the two moments above 
determined must be used for both. 

151. Shear and Moment Curves for Combined Fixed and Mov- 
ing Loads. — In Fig. 50 (PL IV) the line X K- has been taken 
as the axis of abscissas for the curves of shear. Positive shear 
has been laid off downward for fixed loads and upward for 
moving loads. Hence an ordinate at any point drawn perpen- 
dicular to X^ K,, and limited by the two curves (or rather broken 
lines), represents the greatest positive shear due to combined 
fixed and moving loads. The shear curves for fixed and mov- 
ing loads intersect at a point directly opposite the joint N. 
Hence for sections to the right of this point, the shear is 
always negative. 

Similarly, A'"^ Y,n is the axis of abscissas for the two moment 
curves. The bending moment is always negative ; in case of 
the dead load curve, it is laid off upward from X^Kn, while for 
the live load curve it is laid off downward. Hence an ordinate 
between the live and dead load curves at any point of X^^ F„, 
represents the greatest possible bending moment at the cor- 
responding section of the truss. 

We are now prepared to determine maximum stresses in the 
truss members. 

152. Maximum Stresses in Web Members. — The greatest 
stress in any web member is readily found from the shear 
curve. By Art. 139, the stress has such a value that its 
resolved part in the vertical direction is equal to the shear in a 
section through the member. For example, for the member 
14-15 we draw a line in the shear diagram parallel to 14-15 and 
limited by the two lines which determine the greatest shear 



128 GRAPHIC STATICS. 

sustained by the member. This gives 14'-! 5' as the stress in 
the given member. In the same way, I2'-I3', I3'-I4', etc., 
represent stresses in the corresponding truss members. 

As to the sign of the stress in any member, we know that 
for a positive shear any diagonal member has to resist a ten- 
dency of the portion of the truss to the left to move upward ; 
hence the stress is a tension for every member sloping down- 
ward from left to right (as 13-14), and a compression for a 
member sloping downward from right to left (as 14-15). The 
reverse will be true if the shear is negative. We may, how- 
ever, neglect the negative shears given by the curve of Fig. 50, 
because they are the minimum negative shears. (This is evi- 
dent because^ they were determined as maximum positive 
shears ; and since they are found to be really negative, they 
are the least negative shears that can occur at those sections.) 
The maximiifn negative shear for this portion of the truss will 
be found when the train crosses the bridge from left to right ; 
and its value for any section is the same as that of the greatest 
positive shear for the section equally distant from the middle of 
the truss but on the other side. 

For example, since we have found a negative shear in sec- 
tions through the member 21-22, that member sustains a com- 
pression which is represented in the figure by 21 '-22'. But if 
the train is headed to the right, the member 21-22 may sustain 
a much greater compression — equal, in fact, to that already 
found for 14-15, and represented by I4'-I5'. The latter is there- 
fore t\\Q^ greatest compression sustained by 21-22. 

It is evident, therefore, that it is necessary to determine 
stresses from Fig. 50 for those members only which sustain 
positive shear ; remembering that the stress thus found in any 
member is a possible value of the stress in the corresponding 
member on the other side of the center of the truss. 

If both maximum and minimum stresses are desired, it must 
be remembered that the dead-load stresses may occur alone. 

It will be seen that a reversal of stress may occur in the 



PARALLEL CHORDS — CONCENTRATED LOADS. 129 

members 16-17, i/'I^j 18-19, 19-20; while in the remaining 
members no reversal is possible. 

153. Maximum Chord Stresses. — From the maximum bend- 
ing moments represented in Fig. 50, we may readily find the 
greatest stress sustained by any chord member. Consider, 
for example, the member 15-2. If a section be taken cutting 
the three members 7-14, 14-15, and 15-2, and the origin of 
moments be taken at R, it is evident that the sum of the 
moments of the reaction and loads to the left of R is equal to 
the moment of the force acting in the member 15-2. But the 
former is the bending moment in the section through R, and 
its greatest value is equal to the ordinate between the two 
moment curves already drawn, multiplied by the pole distance. 
Hence we have the equation : 

Ordinate of moment curve x pole distance = stress in chord 
member x depth of truss. 

That is, in the present example, the stress in pounds is equal to 
the ordinate multiplied by 120000 and divided by 15. The 
computation may be made graphically as follows : From any 
point Z (Fig. 50) draw two lines making a convenient angle 
with each other, say a right angle. Lay off ZZ" equal to the 
pole distance on the force scale already adopted, and ZZ' equal 
to the depth of the truss, and draw Z'Z" . Then take Z2 equal 
to the ordinate of the moment curve and draw 2-15 parallel to 
Z'Z'\ intersecting ZZ" in the point marked 15. The distance 
Z-i^ represents the stress in 15-2, to the force scale already 
adopted. 

If the pole distance ZZ" had been drawn to some other scale, 
Z-i^ would represent the required stress to the same scale. 

For two chord members symmetrically situated with respect 
to the middle of the truss, the maximum stresses have the same 
value, which is to be computed from the greater of the two 
corresponding values of the bending moment. This is apparent, 
if it is remembered that when the train is headed to the right, 



I30 



GRAPHIC STATICS. 



the two members interchange their conditions. The figure 
shows all the chord stresses. Evidently, all upper chord mem- 
bers are in compression and all lower ones in tension. 

If minimum stresses are desired, they will be found by using 
the dead-load bending moments alone. 

The computation of chord stresses will be facilitated if the 
number of force units in the pole distance is taken as a simple 
multiple of the number of linear units in the depth of the 
truss. For, if H is the pole distance, h the depth of truss, y 
the ordinate of the moment curve, and x the stress in the chord 
member, we have 

Hy = hx', OY x= — y. 
Ji 

M 
Now if — = n, we have only to measure y in linear units, multiply 

by n, and the result is x in force units. 

§ 4. Parallel Chords — Uniformly Distributed Moving Load. 

1 54. Distribution of Load. — In designing highway bridges, 
it is usual to assume the moving load to be uniformly distributed 
along the bridge. The same assumption is sometimes made 
for railroad bridges. This simplifies the computation of stresses 
very materially. In fact, when the chords are parallel, the 
algebraic method of treating such cases becomes so simple that 
it is by many preferred to the graphic method. It will be 
well, however, to indicate the main points in the graphic con- 
struction. 

Maximttm shear. — First it may be noticed that the position 
of moving load for greatest shear in any panel is readily deter- 
mined. The conclusion deduced in Art. 147 may be shown to 
apply to this case ; hence we have the principle that the greatest 
shear in any panel occurs when the portion of the truss to the 
right is completely loaded, while the load on the panel is 
the same fraction of the whole load that the panel length 
is of the whole span. Let A'F(Fig. 51) represent a truss of 



PARALLEL CHORDS — DISTRIBUTED LOAD. 



131 



seven equal panels ; and for the greatest shear in a panel CD 
let ^ be the foremost point of the moving load. Then we must 
have ^Z> equal to one-seventh oi ZY\ or, Zi^ = one-sixth DY. 
A similar rule applies to each panel. 

Maximum moment. — From the principles deduced in Arts. 
129 and 149, it is evident that the greatest bending moment 
occurs at every section when the moving load covers the whole 
bridge. 

Funicular polygon. — In order to draw the funicular polygon, 
it is practically necessary to substitute concentrated loads for 
the uniformly distributed load. In determining shear, we ought 
strictly to have the true funicular polygon for the distributed 
load ; since without it we cannot graphically determine the 
amount of load actually carried at each joint. It will, however, 
be sufficiently correct to subdivide the load into small portions, 
each being assumed concentrated at its center of gravity. 

The student will readily carry out the construction just out- 
lined, every step being similar to the corresponding part of the 
process shown in Fig. 50 (PI. IV). 

155. Assumption of Equal Panel Loads. — The determination 
of stresses, whether graphically or algebraically, is simplified by 
an approximate assumption which will now be explained. This 
assumption is that the moving load on the truss at any instant 
consists of equal loads concentrated at the panel joints ; each 
load being equal to the total load on a length equal to half the 
sum of the two adjacent panels. Thus, in Fig. 51, in computing 
the shear in the panel AB, we assume full panel loads at B, C, 
D, E, and F] and similarly for any other panel. The shear 
will then be equal to the reaction at Xdue to the series of loads 
to the right of the panel considered, and the moment curve will 
be simply a funicular polygon drawn for a series of full panel 
loads at all the joints. 

It will be seen that the error involved in the above assump- 
tion is on the side of safety, so far as the web members are 



132 



GRAPHIC STATICS. 



concerned, since it gives greater values of the shear than 
the more exact method. In case of the bending moments, the 
results are correct for all joints of the loaded chord. For the 
other joints there will be a slight error. This error will be 
avoided, if, in drawing the funicular polygon, the load be con- 
sidered as partly supported at points in the same vertical lines 
with the joints of the unloaded chord; thus, in Fig. 51, equal 
loads should be assumed to act at all the points A', A, B\ B, etc. 
The construction just indicated will not be here shown, being 
very similar to that explained in the following articles for the 
case of a truss with non-parallel chords. 

§ 5. Truss with Ciirved CJiords — Uniform Panel Loads. 

1 56. General Statement. — If the upper and lower chords of 
the truss are not parallel, the determination of the stresses is 
somewhat less simple than in the case of parallel chords. But 
when the live load is taken as uniformly distributed along the 
bridge, and applied to the truss in the way explained in Art. 
155, the graphic construction for finding the maximum stresses 
is not difficult ; and is much less laborious than the algebraic 
computation. 

In Fig. 52 (PI. V) is shown a truss with curved chords, 
divided into equal panels. It will be seen that the method 
given in the following articles applies equally to the case of 
unequal panel lengths. 

The principle of .counterbracing will be employed here, the 
diagonals being constructed to sustain tension only. 

157, Dead Load Stresses. — The dead load stresses maybe 
determined by means of a stress diagram, as in the roof-truss 
problems already treated. Two points should be observed in 
drawing this diagram, (i) If dead loads are taken to act at 
upper as well as at lower joints, the force polygon for the loads 
and reactions must show these forces in the same order as that 
in which their points of application occur in the perimeter of 



CURVED CHORDS — UNIFORM LOADS. 133 

the truss. (2) The diagonal members assumed to act are 
taken as all sloping in the same direction. 

The reason for the first point is the same as already explained 
in Art. 90, — viz., that unless the forces be taken in the order 
mentioned, the stress diagram cannot be the true reciprocal of 
the truss diagram and certain lines will have to be duplicated. 
The reason for assuming the diagonals as all sloping in the 
same way is the same as in the case of the roof truss with 
counterbracing (Arts, iii and 112). 

The dead load stress diagram is not shown, since its con- 
struction involves no principle not already fully explained and 
illustrated. 

158. Chord Stresses Due to Live Load. — It can be easily 
shown that a load at any point of the truss produces tension in 
every lower chord member and compression in every upper 
chord member. Thus, referring to Fig. 52 (PI. V), let a load 
act at^, and let us determine the kind of stress caused in the 
member qq\ Take a section cutting qq\ q'a\ d'd. Consider- 
ing the portion of the truss to the right of the section, the 
only forces acting on it are the reaction at the support and the 
forces in the three members cut. Apply the principle of 
moments to this system of forces, taking the origin at the 
point of intersection of q^d' and dd. It is evident that qq' 
must be in compression to resist the tendency of the reaction 
to produce left-handed rotation about the origin. By similar 
reasoning, assuming a load at any other point, the student will 
be able without difficulty to verify the general statement above 
made. 

It follows that in order to get the greatest possible stresses 
in the chord members, the truss should be fully loaded. 
Hence, to find the live-load chord stresses, a convenient 
method is to draw a stress-diagram for the truss under all live 
loads. This diagram is not shown. 



134 GRAPHIC STATICS. 

159. Live Load Stresses in Web Members. — When the diag- 
onals and verticals are considered, it will be found that loads 
in different positions may cause opposite kinds of stress in any 
member. Thus, considering the member fn\ it is easy to 
show that a tension is caused in it by a load at either of the 
points ab^ be, cd, de, or ef, while a compression is caused by a 
load at fg, gh, or hi. (This may be shown by taking a section 
Xhrough ff ,f^ n\ and n^n, and taking moments about the point of 
intersection of the two chord members cut.) Similarly, loads 
at abj be, ed, de, and ef all tend to throw compression on the 
vertical member f'm', while loads at fg, gh, and /// have the 
opposite tendency. 

Therefore, to produce the greatest tension upon f'n' (and 
compression onf'i?t'), the live load must act only at ef and all 
joints to the right ; while to cause the greatest compression on 
f'n' (and tension upon fin'), the live load must act only at fg, 
gh, and hi. A similar statement will hold regarding any other 
web member. Since counterbraces are to be used in all panels 
in which diagonal members would otherwise be thrown into 
compression, we shall need only to consider the greatest tension 
in each diagonal and the greatest compression in each vertical. 
We shall first outline the method to be employed, and then 
explain the construction. 

To determine the greatest tension in a diagonal member, as 
g'm' : Assume the live load to come upon the bridge from the 
right until there are full loads at the joints ab, be, ed, de, ef and 
fg. Take a section cutting g'm' and the two chord members 
gg' and m'm, and consider the forces acting upon the portion of 
the truss to the left of the section. These forces are four in 
number : the reaction at the support and the forces acting in the 
three members cut. Hence we first determine the reaction, 
and then determine the three other forces for equilibrium by 
the method of Art. 42. 

The construction is shown in Fig. 52 (PI. V). ABCDEFGHI 
is the force polygon for the eight live loads that may come 



CURVED CHORDS — UNIFORM LOADS. 



135 



upon the truss. Choosing a pole O, the funicular polygon 
for the eight loads is next drawn. Now, turning the attention 
to the member ^'?;2', the loads ^/^ and /li are assumed not to act. 
The reactions at the supports for this case of loading are found 
in the usual way. Prolong oa and og to intersect the lines 
of action of the two reactions, and join the two points thus 
determined. This gives the closing line of the funicular poly- 
gon (or om). The ray OM is now drawn parallel to the string 
07/1, and the two reactions are GM a.nd MA. 

Now take a section through vwi' , vi'g' , and g'g, and apply 
the construction of Art. 42 to the determination of the forces 
acting in the three members cut. The resultant of GM and 
the force in gg' must act through the point X (the intersection 
of gg' produced and ij). The resultant of the forces in ;;/;;/' 
and m'g' must act through their intersection Y. Hence these 
two resultants (being in equilibrium with each other) must both 
act in the line XY. From G draw a line parallel to gg' , and 
from M -a. line parallel to XY; mark their point of intersection 
_G' . Then MG' is the resultant of the forces in mm' and m'g' . 
From a^draw a line parallel to mm\ and from G' a line parallel 
to the member g'm', and mark their point of intersection M'. 
Then M' G' represents the force in the member m'g' . This is 
also the value of the greatest stress in m'g'. 

To determine the .stress in the vertical member I'g' , the 
same loading must be assumed, and a similar construction is 
employed. Take a section cutting II' , I'g', and g'g, and deter- 
mine forces acting in these three lines which shall be in equi- 
librium with the reaction GM. This reaction is in equilibrium 
with MG' and G'G, the former having the line of action XY. 
Then MG' is resolved into two forces having the directions of 
the members g' I' and I' I. The stress in g'l' is found to be a 
compression, represented in the stress diagram by the line G'L'. 

The maximum live load stress in every web member may be 
found in the same way. If the above reasoning is understood, 
there will be no difficulty in applying the same method to the 
remaining members. 



136 GRAPHIC STATICS. 

160. Maximum Stresses. — By combining the stresses due to 
live and dead loads, the maximum stresses are easily deter- 
mined. 

Wed members. — When the web members are considered, the 
effect of counterbracing needs careful attention. 

The construction above explained gives the greatest live load 
tension in each diagonal. It may be that for certain members, 
this tension is wholly counterbalanced by the dead loads. In 
any panel in which this is the case, the member shown will 
never act and may be omitted. The counterbrace must then 
be considered. 

Evidently, the algebraic sum of the stresses due to live and 
dead loads will be the true maximum tension in each of the 
diagonals shown. For the greatest tension in the other system 
of diagonals, the load must be brought on the bridge from the 
left ; or, what amounts to the same thing, the tension already 
found for any one of the diagonals shown is also the greatest 
tension in the diagonal sloping the opposite way in the panel 
equally distant from the middle of the truss. (In fact, the 
stresses in all members due to a movement of the loads from 
left to right may be obtained from the results already reached, 
by consideration of symmetry.) 

As to the vertical members, two values of the stress must be 
compared in every case — namely, the greatest compressions 
corresponding to the two directions of the moving load. But 
both can be obtained from the above results by considering the 
symmetry of the truss. For example, the stress found for l^g^ 
is a possible value for the stress in r^b\ and must be compared 
with the value obtained for the latter member when the load 
moves from right to left. It is possible, also, that certain of 
the verticals may be in tension when the dead loads act alone. 

Maximum chord stresses. — These are found by combining 
the stresses due to fixed and moving loads, determined as 
already described. 



CURVED CHORDS — CONCENTRATED LOADS. 137 

161. Example. — The following example should be carefully 
solved, following the method outlined in the preceding articles. 

Given the span, 120 ft. ; lower chord straight ; upper chord 
circular with rise at middle point of 20 ft. ; panel length, 15 
ft. ; diagonals counterbraced to sustain tension only ; width of 
bridge between trusses, 18 ft. Assume dead load from the 
formula of Art. 135, one-third applied at upper chord and two- 
thirds at lower. Live load 85 lbs. per square foot. Determine 
maximum stresses due to live and dead loads. 

162. Parallel Chords. — It will be noticed that all the methods 
which have been described for the discussion of trusses with 
curved chords are applicable to the case of parallel chords. In 
fact, the determination of stresses in web members is simplified 
when both chords are horizontal ; for, after the reaction is 
found as in Art. 159, it is only necessary to so determine the 
stress in the web member that its vertical resolved part shall 
equal the reaction. 

§ 6. Truss ivith Curved Chords — Concentrated Loads. 

163. Comparison of Cases. — When the moving load consists 
of a series of unequal weights, the non-parallelism of the chords 
somewhat complicates the determination of stresses. This case 
can, however, be treated without great difficulty, as will now be 
explained. 

For the determination of chord stresses the method will be 
almost exactly the same as in the case of parallel chords already 
discussed. It is only necessary to determine the greatest bend- 
ing moment at each joint, just as was done in Art. 150, and 
then find the chord stresses by the principle of moments. 

But for the determination of web stresses the shear curve is 
not sufficient, since the shear in any section is not borne wholly 
by the web member, but partly by the inclined chord member. 
Moreover, the position of the moving load which will produce 



138 



GRAPHIC STATICS. 



the greatest shear in a panel is not generally the position which 
causes the greatest stress in the web member in that panel. 

164. Position of Loads for Greatest Stress in a Web Member. 

— We shall now deduce a rule for determining what position of 
the loads will produce the greatest stress in any web member. 
Let XV (Fig. 53) represent the truss and the member con- 




sidered. We know in a general way (Art. 159) that for the 
greatest tension in B^C the truss should be completely loaded 
on the right of the panel BC.^ Let IV = total load on truss ; 
IV' = load on panel BC; I = total length of truss ; /' = length 
of panel BC; x = distance from V to center of gravity of IV; 
x' = distance from C to center of gravity of W. Prolong the 
two chord members of the panel to intersect at Z, and let 
XZ = a, ZB = b. Let R = reaction at X, and P = portion of 
W carried at B. Then 



R = 



IVx 



P = 



W'x 



I ^' 



Now if the truss be separated by a section cutting B' C and the 
two chord members, it is seen that the moment of the stress in 
B'C dihoMt ^ must equal the sum of the moments of R and P 
about Z. Hence the stress in B' C is greatest when the sum 
of the moments of R and P about Z is greatest. Let M = that 
sum, then 

M-. 



Ra-Pt='^x-'^x' 



I 



I' 



* This statement is true for all forms of truss considered in this work. If the two chord 
members in any panel intersect between the vertical lines through the ends of the truss, 
the statement no longer holds. The effect on the web member of a load in any position 
can be determined very easily, whatever the form of the truss, by reasoning similar to that 
employed in Art. 159. 



CURVED CHORDS — CONCENTRATED LOADS. 



139 



If 3/ is a maximum, we must have — - = o. But since dx' = dx, 

dx 

we have, 

dM Wa Wb jrr b I ..„ 

dx I I' a I' 

The ratio - will be known for each panel, and also the ratio —- ; 
a I 

hence the condition expressed by the equation can be easily 
applied. 

It may happen that the condition is satisfied for more than 
one position of the loads. If this is so, such positions must 
all be tried and the results compared. This is illustrated in 
the following article. 

Special case. — It will be noticed that if the chords approach 
parallelism, the point Z moves farther away and the limit 

approached by — is unity. Hence, for the case of parallel 

^ I 

chords, the equation becomes W=-~ W\ This is identical 

with the result already given in Art. 147, 

[Note. — The remarks made in the note in Art. 147 apply also to the reasoning 
just given. It is also to be noticed that we have assumed that no load is between 
B and X. If the condition above deduced cannot be satisfied without carrying the 
foremost load or loads to the left of j5, we may reason thus: Let /F" = load to left 
of B; x" = distance of center of gravity of IF" from B. Then 

M =Ra- Pb- W" (b - x") = ^ a- _ — jr' - W" (b - x"). 

IV ^ ^ 

Differentiating, remembering that dx" — dx' = dx, we have 

dM IVa Wb , „,,, 
dx I I' 

Placing this equal to zero, we have 

a V a 

Usually W" will be zero and the equation first deduced will apply, but if necessary 

the last equation may be used. For the case of parallel chords, the term - IV 

disappears, since - approaches zero.] 
a 



140 



GRAPHIC STATICS. 



165. Determination of Web Stresses. — The method of deter- 
mining web stresses for the case now under consideration is 
shown in Fig. 54 (PL V). The force polygon and funicular 
polygon for the series of wheel loads is drawn in the usual 
manner. The span of the truss here taken is 120 ft., divided 
into eight equal panels, the upper chord being curved, while 
the lower (supporting the floor system and moving loads) is 
straight. Half of the truss is shown at XV, drawn to the 
same scale used in the space diagram for the moving loads. 
The construction is shown for finding stresses in the two mem- 
bers tu and nv. First consider the latter. 

To apply the rule of the preceding article for finding the 

required position of loads, determine the ratio -. Prolonging 

a 

the two chord members k?i and vj7i till they intersect at Z2, we 
find -^- — — = 2. Also we have - = 8. Hence, for the great- 

est stress in tro, the load on the panel QR must equal one- 
sixteenth the load on the whole truss. Now it is easily seen 
that this condition is satisfied when the second load is at the 
point R. For, in this position, the whole load on the truss is 
123750 lbs. If the second load is just at the left of R, the 
load on the panel is 20000 lbs., which is greater than one- 
sixteenth of 123750 lbs. ; but if the second load is just at the 
right of R, the load on the panel is 7500 lbs., which is less than 
one-sixteenth of 123750 lbs. 

We are now ready to determine the stress in the member nv. 
The general method is to cut the truss by a section through 
/'?/, iiv, and V77i ; determine the resultant of all forces on the 
truss to the left of the section ; and then determine three 
forces in the members cut which shall be in equilibrium with 
that resultant. 

The forces acting on the truss to the left of the section are 
the reaction at the support and the portion of the foremost 
load AB carried at the point Q. Their resultant is found as 
follows : Draw X„X„ to represent the span when the load BC'is 



CURVED CHORDS — CONCENTRATED LOADS. 141 

at R ; then draw the closing line of the funicular polygon for 
the forces now on the truss, and from O draw the ray parallel 
to this closing line. The end of this ray is marked Mi, and the 
left reaction is represented by the line MiA. Next replace AB 
by its two components borne at Q and R. These are deter- 
mined by drawing verticals from Q and R intersecting the 
strings oa and od in Q' and R' ; then Q'R' is the string which 
must replace oa and ol?. (The portions of the funicular poly- 
gon to the right of R' and to the left of Q' are not changed by 
thus replacing AB by its two components at Q and R.) Now 
by drawing from O a ray parallel to Q'R' we find the point 
(marked K^) which divides AB into the two components at Q 
and R. 

The resultant of the reaction at X and the load at Q is given 
in magnitude by AfiKi. We need also its line of action, which 
is found as follows : Prolong Q'R' and omi (the closing string of 
the funicular polygon for all the forces on the truss) until they 
intersect in s' ; the required line of action is a vertical line 
through 2'. This vertical line intersects X^X^ produced in s. 
We now locate the point -s- in the truss diagram by making 
X.:; = XuZ in the diagram above. 

We have now to solve the following problem : Determine 
three forces acting in lines kit, uv, and vm, which shall be in 
equilibrium with a force equal to the resultant just found and 
acting upward in a line through z. This is solved by the 
method of Art. 42. Draw KM equal to the given force acting 
at z. Draw zQ', and determine two forces, one parallel to 
zQ' and the other to hi, which shall be in equilibrium with 
KM. This gives UK as the force in the line tik and MU as 
the resultant of the forces in the lines mv and vii. From M 
and U draw lines parallel to niv and vil respectively, and mark 
their intersection V\ then J/Fand F^" represent the stresses 
in the corresponding truss members. The latter is evidently 
the required stress in the member tiv. 

Turning next to the member /;/, a section through it will cut 



142 



GRAPHIC STATICS. 



the two chord members kiL and mt ; these intersect at Z^. By 

measurement we find -=:^^^=i6, and --- = 8x6 = 48. There- 

a Z^X I a 

fore, for the greatest stress in tii we must have the load on the 
panel QR equal to the whole load on the truss divided by 48. 
This condition is seen to be satisfied when the foremost load is 
at R. Since in this case there is no load borne to the left of 
R, the only force acting on the truss to the left of a section 
through ttL is the reaction at the support. To find this, draw 
XtXu to represent the length of the truss for the required posi- 
tion of the loads, and find oifix, the closing string of the funic- 
ular polygon for all loads and reactions on the truss. The 
corresponding ray is OM^, and M^A represents the reaction 
at the left support. 

Referring now to the truss diagram, we have to determine 
forces acting in the three lines int^ tiiy tik, which shall be in 
equilibrium with the reaction just determined. Lay off K^Mto 
represent this reaction ; draw a line from K' parallel to he and 
one from M parallel to QX, and mark their point of intersec- 
tion U' ; then U'K^ represents the force in tik, and MU' the 
resultant of the forces in mt and ///. From M draw a line par- 
allel to mt and from U^ a line parallel to tit, intersecting at T] 
then TU^ represents the required stress in tu. 

The value just determined is the true maximum stress for 
the member tn. In case of the member iro, however, farther 
investigation is needed. It will be found by trial that the 

condition W = ~ - ■— • W is satisfied for the member itv, not 
a I' 

only when the loads have the position already treated, but also 
when the foremost load is at R. Hence the stress in uv for 
this position of the loads must be determined and compared 
with the one already found, namely UV. For this case of 
loading the only force acting on the truss to the left of a section 
through tiv is the reaction (already found when considering the 
member /;/) represented by K' M. Making this in equilibrium 
with three forces whose lines of action are kii, ?iv, vni, we find 



CURVED CHORDS — CONCENTRATED LOADS. 



143 



U^ V as the stress in uv. Comparison shows that this is 
sUghtly greater than L^V; hence the value last found is the 
true maximum for the member tiv. 

The construction for each of the other web members is 
exactly similar. If both chord members in any panel are 
inclined, the construction requires only slight modification, 
which the student will readily supply. 

In case the intersection of two chord members cannot con- 
veniently be found by prolonging them, the distances a and /; 
can be easily computed from the dimensions of the truss. 

Remark. — It will often be found that two web members 
related as are tii and tiv, will not sustain their maximum stresses 
at the same time. In the case of parallel chords this could not 
occur. 

166. Minimum Stresses. — In what precedes, little has been 
said of minimum stresses in the truss members. If, however, 
the design is to be made in accordance with the theory of the 
strength of materials under repeated alternations of stress, the 
minimum stress sustained by each member becomes important. 
In all the cases treated in the present chapter, the minimum 
stresses can be determined without difficulty, without the use 
of additional principles. 



Part III. 



CENTROIDS AND MOMENTS OF INERTIA. 



:)>a<c 



CHAPTER VIII. CENTROIDS. 



§ I. Centroid of Parallel Forces. 



167. Composition of Parallel Forces. — The composition of 
complanar parallel forces can always be effected by means 
of the funicular polygon, by the method of Art. 27. It is now 
necessary to consider parallel systems more at length, as a pre- 
liminary to the discussion of graphic methods for determining 
centers of gravity and moments of inertia. 

168. Resultant of Two Parallel Forces. — Let ab and be (Fig. 
55) be the lines of action of two parallel forces, their magni- 
tudes being AB and BC (not shown). Let 
ac be the line of action of their resultant, 
and AC (not shown) its magnitude. By 
the principle of moments (Art. 50) the sum 
of the moments of AB and BC about any 
point in their plane is equal to the moment 

of ^6' about the same point. If the origin of moments is on 
aCy the moment of AC is zero ; and therefore the moments of 
AB and BC are numerically equal (but of opposite signs). 

Let any line be drawn perpendicular to the given forces, 
intersecting their lines of action in P', Q', and R' respectively. 
Then ABxP'P' = BCx Q'R'. 



R 



R' 



Q' 



Q 



"Fiff. £55 



144 



CENTROID OF PARALLEL FORCES. 



145 



Let any other line be drawn intersecting the three Hnes of 
action in P, Q, and R respectively. Then 

PR ^ QR 
P'R' Q'R'' 

and therefore ABxPR = BCx QR. 

That is, PQ is divided by the line ac into segments inversely 
proportional to AB and BC. 

If AB and BC act in opposite directions, the line ac will be 
outside the space included between ab and be ; but the above 
result is true for either case. 

169. Centroid of Two Parallel Forces. — If the lines of action 
ab and be (Fig. 55) be turned through any angle about the points 
P and Q respectively, the forces remaining parallel and of 
unchanged magnitudes, the line of action of their resultant will 
always pass through the point R. For, by the preceding article, 
the line of action of the resultant will always intersect PQ in a 
point which divides PQ into segments inversely proportional to 
AB and BC. Hence, if AB and BC remain unchanged, and 
also the points P and Q, the point R must also remain fixed. 

If P and Q are taken as the points of application of AB and 
BC, R rnay be taken as the point of application of AC, in 
whatever direction the parallel forces are supposed to act. The 
point R is called the centroid * of the parallel forces AB and 
BC for the fixed points of application P and Q. 

170. Centroid of Any Number of Parallel Forces. — Let AB, 

BC, and CD be three parallel forces, and let P, Q, and 5 
(Fig. 56) be their fixed points of 
application. Let R be the cen- 
troid of AB and BC, and ^4 (7 their 
resultant. Take R as the fixed 
point of application of AC, and 
determine T, the centroid oi AC 

* The name center of parallel forces has been quite commonly used instead of centi'oid 
as above defined. The latter term has, however, been adopted by some of the later writers, 
and seems on the whole a better designation. 




146 



GRAPHIC STATICS. 



and CD. Let AD be the resultant of AC d.nd CD; then AD is 
also the resultant of AB, BC, and CD. 

Now if ABy BC, and CD have their direction changed, but 
still remain parallel and unchanged in magnitude, it is evident 
that the point T, determined as above, will remain fixed and will 
always be on ad, the line of action of AD. The point Tis called 
the centroid of the three forces AB, BC, and CD. 

By an extension of the same method, a centroid may be 
determined for any system of parallel forces having fixed points 
of application. Hence the following definition may be stated : 

The centroid of a system of parallel forces having fixed points 
of application is a point through which the line of action of 
their resultant passes, in whatever direction the forces be taken 
to act. 

In determining the centroid by the method just described, 
the forces may be taken in any order without changing the 
results. For the centroid must lie on the line of action of the 
resultant ; and since this is a determinate line for each direction 
in which the forces may be taken to act, there can be but one 
centroid. 

171. Non-complanar Parallel Forces. — The reasoning of the 
preceding articles is equally true, whether the forces are com- 
planar or not. In what follows we shall deal either with 
complanar forces, or with forces whose points of application are 
complanar. No more general case will be discussed. 

172. Graphic Determination of Centroid of Parallel Forces. — 

If the line of action of the resultant of any system of parallel 
forces be found for each of two assumed directions of the forces, 
the point of intersection of these two lines is the centroid of 
the system. Moreover, if the points of application are com- 
planar, the two assumed directions may both be such that the 
forces will be complanar. 

Thus, let the plane of the paper be the plane containing the 
given points of application, and let ab, be, cd, de, ef (Fig. 57) be 



CENTROID OF PARALLEL FORCES. 



147 



the points of application of five parallel forces, AB, BC, CD, 
DE, EF. Draw through these points parallel lines in some 
chosen direction, and taking them as the lines of action of the 
given forces, construct the force and funicular polygons corre- 




sponding. The line of action of the resultant is drawn through 
the intersection of the strings oa and of, and this line contains 
the centroid of the given forces. Next draw through the given 
points of application another set of parallel lines, preferably per- 
pendicular to the set first drawn, and draw a funicular polygon 
for the given forces with these lines of action. (It is unneces- 
sary to draw a new force diagram, since the strings in the 
second funicular polygon may be drawn respectively perpendic- 
ular to those of the first.) This construction determines a 
second line as the line of action of the resultant corresponding 
to the second direction of the forces. The required centroid of 
the system is the point of intersection of the two lines of 
action of the resultant thus determined, and is the point af in 
the figure. 

Example. — Find graphically the centroid of the following 
system of parallel forces, and test the result by algebraic com- 
putation : A force of 20 lbs. applied at a point whose rectangu- 
lar coordinates are (4, 6) ; 12 lbs. at the point (12, 3) ; 20 lbs. at 



148 



GRAPHIC STATICS. 



the point (lo, lo) ; — lo lbs. at the point (7, —9); —8 lbs. at 
the point (—5, —10). 

173. Centroid of a Couple. — If AB and BC are the magni- 
tudes of any two parallel forces applied at points P and Q, 
then by Art. 169 their centroid R is on the line PQ and is 
determined by the equation 

PR^BC 
QR AB' 

If AB and BC are equal and opposite forces, PR and QR must 
be numerically equal, and R must be outside the space between 
the lines of action of AB and BC. These conditions can be 
satisfied only by making PR and QR infinite. Hence we may 
say that the centroid of two equal and opposite forces lies on 
the line joining their points of application and is infinitely dis- 
tant from these points. 

174. Centroid of a System whose Resultant is a Couple. — 

If a system with fixed points of application is equivalent to a 
couple, its centroid will be infinitely distant from the given 
points of application. A line containing this centroid can be 
determined as follows : 

Take the given forces in two groups ; the resultants of the 
two groups will be equal and opposite. Find the centroid of 
each group, and suppose each partial resultant applied at the 
corresponding centroid. Then the centroid of the whole sys- 
tem is the same as that of the couple thus formed, and will lie 
on the line joining the two partial centroids. 

If the separation into groups be made in different ways, dif- 
ferent couples and different partial centroids will be found. 
The different couples are, of course, equivalent ; and it may be 
proved that the lines joining the different pairs of partial cen- 
troids are all parallel, and intersect in the (infinitely distant) 
centroid of the whole system. 

For, suppose the given system to be equivalent to a couple 



CENTER OF GRAVITY. 1 49 

Q with points of application A and B, and also to a couple Q' 
with points of application A' and B'. These two couples must 
be equivalent to each other, whatever be the direction of the 
forces. Let AB be taken as this direction. The two equal 
and opposite forces of the couple Q neutralize each other, since 
their lines of action are coincident ; hence the two forces of 
the couple Q' must also neutralize each other. Therefore A'B' 
must be parallel to AB. 

175. Moment of a Force about an Axis. — The moment of a 
force with respect to a given axis, as defined in Art. 47, 
depends not only upon the point of application of the force, 
but upon its direction. In dealing with systems of forces 
whose direction may change but whose points of application 
are complanar, we shall need to compute moments only for 
axes lying in the plane of the points of application ; and the 
forces may usually be regarded as acting in lines perpendicular 
to this plane. Hence we shall compute moments by the 
following rule : 

The moment of a force with reference to any axis is the 
product of the magnitude of the force into the distance of its 
point of application from the axis. 



§ 2. Center of Gravity — Definitions and General Principles. 

176. Center of Gravity of Any Body. — Every particle of 
a terrestrial body is attracted by the earth with a force propor- 
tional directly to the mass of the particle. The resultant of 
such forces upon all the particles of a body is its zveigJit ; and 
the point of application of this resultant is called the center of 
gravity of the body. The lines of action of these forces may 
be assumed parallel without appreciable error. We may there- 
fore define the center of gravity of a body as follows : 

If forces be supposed to act in the same direction upon all 
particles of a body, each force being proportional to the mass 



ISO 



GRAPHIC STATICS. 



of the particle upon which it acts, the centroid of this system 
of parallel forces is the center of gravity of the body. 

This point is also called center of mass, and center of inertia^ 
either of which is a better designation than center of gravity. 
The latter term is, however, in more general use. 

177. Centers of Gravity of Areas and Lines. — The term 
center of gravity, as above defined, has no meaning when 
applied to lines and areas, since these magnitudes have no 
mass, and hence are not acted upon by the force of gravity. 
It is, however, common to use the name center of gravity in 
the case of lines and areas, with meanings which may be stated 
as follows : 

The center of gravity of an area is the center of gravity of 
its mass, on the supposition that each superficial element has a 
mass proportional to its area. This point would be better 
described as the center of area. 

The center of gravity of a line is. the center of gravity of its 
mass, on the assumption that each linear element has a mass 
proportional to its length. The term center of length is pref- 
erable, and will often be used in what follows. 

Similar statements might be made regarding geometrical 
solids, but we shall have to deal chiefly with lines and areas. 

178. Moments of Areas and Lines. — Definition. — The mo- 
ment of a plane area with reference to an axis lying in its plane 
is the product of the area by the distance of its center of 
gravity from the axis. 

Proposition. — The moment of any area about a given axis is 
equal to the sum of the moments of any set of partial areas 
into which it may be divided. For, by the definition of center 
of gravity, a force numerically equal to the total area and 
applied at its center of gravity is the resultant of a system of 
forces numerically equal to the partial areas and applied at 
their respective centers of gravity ; and the moment of any 



CENTER OF GRAVITY. 131 

force is equal to the sum of the moments of its components 
(Art. 50). 

A similar definition and proposition may be stated regarding 
lines. 

The moment of an area or line is zero for any axis containing 
its center of gravity. 

179. Symmetry. — Two points are symmetrically situated 
with respect to a third point if the line joining them is bisected 
by that point. 

Two points are symmetrically situated with respect to a line 
or plane when the line joining them is perpendicular to the 
given line or plane and bisected by it. 

A body is symmetrical with respect to a point, a line, or a 
plane, if for every point in the body there is another such that 
the two are symmetrically situated with respect to the given 
point, line, or plane. The point, line, or plane is in this case 
called a point of symmetry, an axis of symmetry, or a plane of 
symmetry of the body. 

180. General Principles. — (i) The center of gravity of two 
masses taken together is on the line joining the centers of 
gravity of the separate masses. For it is the point of applica- 
tion of the resultant of two parallel forces applied at those 
points. 

{2) If a body of uniform density has a plane of symmetry, 
the center of gravity lies in this plane. If there is an axis of 
symmetry, the center of gravity lies in this axis. If the body 
is symmetrical with respect to a point, that point is the center 
of gravity. For the elementary portions of the body may be 
taken in pairs such that for each pair the center of gravity is 
in the plane, axis, or point of symmetry. 

181. Centroid. — The center of gravity of any body or geo- 
metrical magnitude is by definition the same as the centroid of 
a certain system of parallel forces. It will be convenient, 



152 GRAPHIC STATICS. 

therefore, to use the word centroid in most cases instead of 
center of gravity. 



§ 3. Centroids of Lines and of Airas. 

182. General Method of Finding Centroid. — The centroid 
of any area may be found by the following method : Divide the 
given area into parts such that the area and centroid of each 
part are known. Take the centroids of the partial areas as the 
points of application of forces proportional respectively to those 
areas. The centroid of this system of forces is the centroid of 
the total area, and may be found by the method of Art. 172, 

The centroid of a line may be found by a similar method. 

The method just described is exact if the magnitudes of the 
partial areas and their centroids are accurately known. If the 
given area is such that it cannot be divided into known parts, it 
will still be possible to get an approximate result by this 
method. 

In applying this general method, it is frequently necessary to 
know the centroids of certain geometrical lines and figures, and 
also the relative magnitudes of the areas of such figures. 
Methods of determining these will be given in the following 
articles. 

183. Centroids of Lines. — The centroid of 2. st7'aight line is 
at its middle point. 

Broken line. — The centroid of a broken line is the center of 
a system of parallel forces, of magnitudes proportional to the 
lengths of the straight portions of the line, and applied respec- 
tively at their middle points. It may be found graphically by 
the method of Art. 172, or by any other method applicable to 
parallel forces. 

Part of regular polygoji. ■. — For the centroid of a part of a 
regular polygon, a special construction is found useful. 



CEXTROIDS OF LINES AND OF AREAS. 153 

Let ABCDE (Fig. 58) be part of such a polygon, and O the 

center of the inscribed circle. Let 

r> S-^'^'cT^T — 5 ^ = radius of inscribed circle ; / = 

f^/^^-4^ icl\2y.^_^E length of a side of the polygon ; i- 

I 1 \ ! \\ ! = total length of the broken line 

I i \ 1 \^J 

1^ ! ^xi N\ AE. Through O draw OC, the axis 

^ ^^ ^ ^^^ of symmetry of AE; and MN, per- 

pendicular to OC. 

First, the centroid must lie on OC. 

Second, to find its distance from O, assume a system of equal 
and parallel forces applied at the middle points of the sides AB, 
BCy etc. The required centroid is the point of application of 
the resultant of these forces. Taking MN as the axis of 
moments, and letting x = required distance of centroid from 
MN, and x^, x^, x^, x^ the distances of the middle points of AB, 
BC, etc., from MN, we have from the principle of moments, 

Ixi 4- 1x2 + ^-^3 + ^^i = ^^' 

But if Ab and Bb be drawn perpendicular respectively to PQ 
and MNwe have from the similar triangles ABb, POQ. 

AB^PO J^^r 
Ab~PQ^^ Ab~xi 

,♦. /xi=r ' Ab. 
In the same way, 

1x2 = r ' be, 

1x2, = r ' (^d, 

Ix^ = r • dE. 

Hence, S'X = r {Ab-\-bc-\-cd-\-dE) = r'AE, 

where AE is equal to the projection of the broken line ABCDE 
on MN. 

The centroid G may now be found graphically as follows : 
U3kQ0c' = r; ON=^s', OE' = ^AE; draw iV^'. Then 6^ is 
determined by drawing E' G parallel to Nc'. 



154 



GRAPHIC STATICS. 



Circtdar arc. — The above construction holds, whatever the 
length of the side /. If this length be decreased indefinitely, 
while the number of sides is increased indefinitely, so that the 
length s remains finite, we reach as the limiting case a circular 
arc. The same construction therefore applies to the determina- 
tion of the centroid of such an arc, r denoting the radius of the 
circle and s the length of the arc. 

184. Centroids of Geometrical Areas. — Pa^'allelogram. — The 
centroid of a parallelogram is on a line bisecting two opposite 
sides. 

Let ABCD (Fig. 59) be a parallelogram, and EF a line 
bisecting AD and BC. Divide AB into any even number of 

equal parts, and through the points 

B^ ^^ P — ,_, — C of division draw lines parallel to BC. 

Also divide BCmX-o any even number 



\ \ 




\\ p V \ \Q \ of equal parts and draw through the 

\\ \ \\ \ points of division lines parallel to 

E D 

AB. The given parallelogram is thus 
divided into equal elements. Now 
consider a pair of these elements, such as those marked P and 
Q in the figure, equally distant from AD, and also equally 
distant from EF, but on opposite sides of it. The centroid of 
the two elements taken together is at the middle point of the 
line joining their separate centroids. If the number of divisions 
of AB and of BC be increased without limit, the elements 
approach zero in area, and the centroids of P and Q evidently 
approach two points which are equally distant from EF. Hence 
in the limit, the centroid of such a pair of elements lies on the 
line EF. But the whole area ABCD is made up of such pairs ; 
hence the centroid of the whole area is on the line EF. For 
like reasons it is also on the line bisecting AB and DC\ hence 
it is at the intersection of the two bisectors. 

The point thus determined evidently coincides with the point 
of intersection of the diagonals AC 3.nd BD. 



CENTROIDS OF LINES AND OF AREAS. 



155 




Triangle. — The centroid of a triangle lies on a line drawn 
from any vertex to the middle of the opposite side ; and is, 
therefore, the point of intersection of the three such lines. 

Let ABC (Fig. 60) be any triangle, and D the middle point 
of BC. Then the centroid of ABC must lie on AD. For AD 
bisects all lines, such as be, parallel to BC Now inscribe in 
the triangle any number of parallelograms such as bec^b\ with 
sides parallel respectively to BC and 
AD. The centroid of each parallelo- 
gram lies on AD, and, therefore, so 
also does the centroid of the whole 
area composed of such parallelograms. 
If the number of such parallelograms 
be increased without limit, the alti- 
tude of each being diminished without limit, their combined 
area will approach that of the triangle, and the centroid of this 
area will approach in position that of the triangle. But since 
the former point is always on the line AD, its limiting position 
must be on that line. Therefore the line AD contains the cen- 
troid of the triangle. 

By the same reasoning, it follows that the centroid of ABC 
must lie on BE, drawn from B to the middle point of AC. 
Hence it must be the point of intersection of AD and BE, 
which point must also lie on the line CF drawn from C to the 
middle point of AB. 

The point G divides each bisector into segments which are 
to each other as i to 2. For, from the similar triangles ABC, 
EDC, since EC is half of AC, it follows that DE is equal to 
half of BA. And from the similar triangles AGB, DGE, since 
DE is half of AB, it follows that GE is half of GB, and GD 
half of GA. 

Quadrilateral. — Let ABCD (Fig. 61) be a quadrilateral of 
which it is required to find the center of gravity. Draw BD, 
and let E be its middle point. Make EGi = \ EA, and EG2=\ 
EC Then the centroids of the triangles ABD and BCD are 



156 



GRAPHIC STATICS. 




Fig. 61 



Gy and Gi respectively. Hence the centroid of ABCD is on 
the line G^Gi at a point dividing it into segments inversely 

proportional to the areas of ABD 
and BCD. Since these two tri- 
angles have a common base, their 
areas are proportional to their alti- 
tudes measured from this base. 
But these altitudes are propor- 
tional to AF and FC, or to G^H and G^H \ hence, if G is the 
required centroid, 

G\G '. G-iG : : G-iH : G\H. 

Therefore G is found by making GxG=G<iH. 

Circular sector. — To find the centroid of a circular sector 
OAB (Fig. 62) we may reason as follows : Draw two radii OM, 
ONj very near together. Then OMN differs little from a tri- 
angle, and its centroid will fall very near the arc A^B\ drawn 
with radius equal to two-thirds of OA. If the whole sector be 
subdivided into elements such as OMN, their centers of gravity 

will all fall very near to the arc AB\ If the 
number of such elements is indefinitely in- 
creased, the line joining their centroids 
approaches as a limit the arc A^B\' And 
since the areas of the elements are propor- 
tional to the lengths of the corresponding 
portions of A^B\ the centroid of the total 
area is the same as that of the arc AB' . 
This point may be found by the method described in Art. 183. 

185. Graphic Determination of Areas. — Let there be given 
any number of geometrical figures, and let it be required to 
determine the relative magnitudes of their areas. 

If a number of rectangles can be found, of areas equal 
respectively to the areas of the given figures and having one 
common side, then the remaining sides will be proportional to 
the areas of the given figures. 



M N 




CEXTROIDS OF LIXES AND OF AREAS. 



157 



An important case is that in which the given figures are 
such that the area of each is equal to the product of two known 
Hues. In this case a series of equivalent rectangles having one 
common side can be found by the following construction. • 

Let ABC (Fig. 63) be a triangle, and let it be required to 
determine an equivalent 
rectangle having a side of 
given length as LM. Let 
b and h be the base and 
altitude of the triangle. 
Make ZyV=i/^ and ZP-//, 
and draw the semicircumference PRIV. 
perpendicular to PN] then 
r2 




Draw the ordinate LR 



LR' 



PL X LN=\ bh = 2irt2i ABC. 



Draw AIR, and perpendicular to it draw RQ. Then 

LR^ = LMy.LQ = 2.x^2. ABC 

Hence LQ is the required length. 

If the given figure is a parallelogram, LN and LP may be 
its base and altitude. If it be a circular sector, LN and LP 
may be the length of the arc and half the radius. 

186. Centroids of Partial Areas. — It may be required to 
find the centroid of the part remaining after deducting known 
parts from a given area. For this case the construction of 
Art. 182 needs modification. In the case there considered we 
regarded the partial areas and the total area as proportional to 
parallel forces acting at their respective centers of gravity. In 
this case we may also represent the total area, the portions 
deducted from it, and the remaining portion as forces acting at 
the respective centers of gravity ; but the forces corresponding 
to the areas deducted must be taken as acting in the opposite 
direction to that assumed for the forces representing the total 
area and the area remaining. 

Thus, to find the centroid of the area remaining after deduct- 



158 



GRAPHIC STATICS. 




ing from the circle ABD (Fig. 64) a smaller circle EFH and a 
sector OAB, we may proceed as follows : Find the centroid of 

three parallel forces proportional to the 
areas ABD, EFH, and OAB, and applied 
at their respective centroids O, C, and G ; 
but the last two must be taken as acting 
in the direction opposite to that of the first. 
With this understanding, the force and 
funicular polygons may be employed as in 
JB'is. 64: Art. 172. 

187. Moments of Areas. — The moment of an area about any 
line in its plane may be determined from the funicular polygon 
employed in finding its center of gravity. Let the parallel 
forces applied at the centroids of the partial areas be assumed 
to act parallel to the axis of moments. Then the distance inter- 
cepted on the axis by the extreme lines of the funicular poly- 
gon, multiplied by the pole distance, is equal to the moment of 
the total area about the axis. For, by Art. 56, this construc- 
tion gives the moment of the resultant force about any point 
in the given axis ; and this is equal to the moment of the result- 
ant area about the axis by definition. 

A similar rule gives the moments of the partial areas. 



CHAPTER IX. MOMENTS OF INERTIA. 

§ I. Moments of Inertia of Forces. 

1 88. Definitions. — The moment of inertia of a body with 
respect to an axis is the sum of the products obtained by mul- 
tiplying the mass of every elementary portion of the body by 
the square of its distance from the given axis. 

The moment of inertia of an area with respect to an axis is 
the sum of the products obtained by multiplying each element- 
ary area by the square of its distance from the axis. 

The moment of inertia of a line may be similarly defined , 
using elements of length instead of elements of mass or area. 

The moment of inertia of a foixe with respect to any axis is 
the product of the magnitude of the force into the square of the 
distance of its point of application from the axis. The sum of 
such products for any system of parallel forces is the moment 
of inertia of the system with respect to the given axis. 

[Note. — The term moment of inertia had reference originally to material bodies, 
the quantity thus designated having especial significance in dynamical problems relat- 
ing to the rotation of rigid bodies. The quantity above defined as the moment of 
inertia of an area is of frequent occurrence in the discussion of beams, columns, and 
shafts in the mechanics of materials. In the graphic discussion of moments of iner- 
tia of areas, it is convenient to treat areas as forces, just as in the determination of 
centers of gravity; it is therefore convenient to use the term moment of inertia of a 
force in the sense above defined. It is only in the case of masses that the term 
moment of inertia is really appropriate, but it is by analogy convenient to apply it 
to the other cases.] 

Th^ product of inertia of a mass with respect to two planes is 
the sum of the products obtained by multiplying each element- 
ary mass by the product of its distances from the two planes. 

159 



l5o GRAPHIC STATICS. 

The product of inertia of an area, a line, or a force may be 
defined in a similar manner. 

For a plane area, the product of inertia with respect to 
two lines in its plane may be defined ' as the sum of the 
products obtained by multiplying each element of area by the 
product of its distances from the given lines. This is equiv- 
alent to the product of inertia of the area with respect to two 
planes perpendicular to the area and containing the two given 
lines. 

For a system of forces with points of application in the same 
plane, the product of inertia with reference to two axes in that 
plane may be defined as the sum of the products obtained by 
multiplying the magnitude of each force by the product of the 
distances of its point of application from the given axes. It is 
with such systems and with plane areas that the fpllowing pages 
chiefly deal. 

The radius of gyration of a body with respect to an axis is the 
distance from the axis of a point at which, if the whole mass of 
the body were concentrated, its moment of inertia would be 
unchanged. The square of the radius of gyration is equal to 
the quotient obtained by dividing the moment of inertia of the 
body by its mass. 

The radius of gyration of an area may be defined in a similar 
manner. 

The radius of gyration of a system of parallel forces is 
the distance from the axis of the point at which a force 
equal in magnitude to their resultant must act in order that 
its moment of inertia may be the same as that of the system. 
The square of the radius of gyration may be found by dividing 
the moment of inertia of the system by the resultant of the 
forces. 

189. Algebraic Expressions for Moment and Product of In- 
ertia. — Mome7tt of inertia. — Let ?«i, m<i, etc., represent ele- 
mentary masses of a body, and ?'i, r^, etc., their respective 



MOMENTS OF INERTIA OF FORCES. i6i 

distances from an axis ; then the moment of inertia of the body 
with respect to that axis is 

the symbol X being a sign of summation, and the second mem- 
ber of the equation being merely an abbreviated expression for 
the first. 

Product of inertia. — Let/], /2, etc., denote the perpendicular 
distances of elements ;;^i, m^, etc., from one plane, and q^, q.,, etc., 
their distances from another ; then the product of inertia of the 
body with respect to the two planes is 

inipiqi + m-2p-iq-2 H = %mpq. 

Radius of gyration. — With the same notation, if k denotes 
the radius of gyration of the body, we have 

{nti + ?/?9 H ) k^ = niiVx + nicyV.^- -\ . 

Or, 

72 _ m^r^^ + ^2^2' H — _ ^ w^^ 
m^ -\-in2-\ 2!;;/ 

Here Xm is equal to the whole mass of the body. 

Product-radiiis. — Let c represent a quantity defined by the 
equation. 

{mi + ;;/2 H ) ^ = i^t\p\qi + ni^p-iq^ H • 

Then if the whole mass were concentrated at the same distance 
c from two axes, its product of inertia with respect to those 
axes would be unchanged. This quantity c is thus seen to be 
analogous to the radius of gyration. It may be called the 
product-radius of the body with respect to the two axes. The 
value of c is always given by the equation 

■2_^inpq 

Expressions similar to those just given apply also to plane 
areas and to systems of parallel forces. In case of an area, 
mi^ in2, etc., denote elements of area; Ti, r2, etc., their distances 
from the axis of inertia ; and {pi, qi), {pi, q.^, etc., their dis- 



l62 GRAPHIC STATICS. 

tances from the two planes. In case of a system of parallel 
forces with complanar points of application, nix, ^^2> etc., must 
be replaced by the magnitudes of the forces. 

190. Determination of Moment of Inertia of a System of 
Parallel Forces. — Let the points of application of the forces 
be in the same plane, which also contains the assumed axis. 
We shall have to deal only with systems satisfying these condi- 
tions. By the definition, the moment of inertia will be the 
same, whatever the direction of the forces. If we take the 
moment of any force (as defined in Art. 175) about the given 
axis, and suppose a force equal in magnitude to this moment 
to act at the point of application of the original force, and in a 
direction corresponding to the sign of the moment, then the 
moment of this new force about the given axis is equal to 
the moment of inertia of the original force. If this be done for 
all the forces, the algebraic sum of the results will be the 
required moment of inertia of the system. 

This process can be carried out graphically by methods 
already described. 

Let ab, be, cd, de (Fig. 65) be the points of application of 
four parallel forces, and let the axis of inertia be QR. Suppose 
all the forces to act in lines parallel to QR, passing through the 
given points of application. Their respective moments with 
reference to QR may now be found by the method of Art, 55. 

Draw the force polygon ABCDE and choose a pole O, taking 
the pole distance H preferably equal to AE or some simple 
multiple of AE. (In Fig. 65 H is taken equal to AE) Draw 
a funicular polygon and prolong each string to intersect QR. 
Then the moment of any force with respect to QR is the 
product of H by the distance intercepted on QR by the two 
strings corresponding to the force in question. Thus the 
moment of AB is given by the distance A^ B^ (Fig. 65) multi- 
plied by H. Also the successive moments of BC, CD, DE are 
represented by B^C\ C'D\ D'E', each multiplied by H. It is 



MOMENTS OF INERTIA OF FORCES. 



163 



seen also that the intercepts, if read in the above order, give a 
distinction between positive and negative moments ; upward 
distances on QR denoting in this case positive moments, and 
downward distances negative moments. 

We have now to find the sum of the moments of a second 
system of forces acting in the original lines, but represented in 





inig.65 



magnitude and direction by the intercepts just found. We 
may take as the force .polygon for the second Construction the 
line A^B' C'D'E' (Fig. 65), and choosing any pole distance H', 
draw a second funicular polygon and find the distances inter- 
cepted by the successive strings on the line QR. But in this 
case, since only the resultant is desired, we need only find the 
intercept A'^E" between the first and last strings. The product 
of this intercept by H' gives the sum of the moments of A'B', 
B'C, CD', D'E' with respect to QR \ and, if the product be 
multiphed by H (since A'B', B'C, etc., should each be multi- 
plied by ^in order to represent the magnitudes of the forces 
of the second system), the result will be the required moment of 
inertia of the given system of forces AB, BC, CD, DE. 



164 GRAPHIC STATICS. 

It should be noticed that in Fig. 65 (A) is a /o7re diagrain^ 
{B) a space diagram (Art. 11); that is, every line in {A) repre- 
sents a force, while every line in {B) represents a distance. 
Even A' B' , B'C, etc., though used as forces, are actually 
merely distances ; and the moment of any one of them is the 
product of a length by a length. 

191. Radius of Gyration. — The moment of inertia of the 
given system is HxH' xA"E". If H has been taken equal 
to AB, the product H'xA'^E" must equal the square of the 
radius of gyration of the system with respect to QR. The 
length of the radius of gyration can be found as follows (Fig. 
65) : Draw LN=H' + A"E'\ and make LM=H'. On Z/V^as 
a diameter draw a semicircle LPN, and from M draw a line 
perpendicular to LN^, intersecting the semicircle in P. Then 

MP is the length of the required radius of gyration. For by 

2 

elementary geometry we have PM —LMxMN. 

If H is taken equal to n xAE, the moment of inertia is equal 

to H^ xA"E" X n X AE, and the square of the radius of gyration 

is equal to nH' xA" E". Hence in Fig. 65 we should put 

either LM=nH', or MN=nxA"E". 

192. Central Axis. — If the axis with reference to which the 
moment of inertia is found contains the centroid of the given 
system of forces, it is called a central axis of the system. 

In many cases it is desired to find the moment of inertia with 
respect to a central axis whose direction is known while the 
position of the centroid is at first unknown. It is to be noticed 
that the method shown in Fig. 65 is applicable in this case ; 
for the first part of the process is identical with that employed 
in finding the centroid of the system. If, in Fig. 65, the 
strings oa^ oe, of the first funicular polygon be prolonged to 
intersect, a line through their point of intersection, parallel to 
the direction of the forces, will contain the centroid of the 
system. If this line is taken as the inertia-axis, the points A', 
B' , C , D\ E' are the points in which this axis is intersected by 



xMOMENTS OF INERTIA OF FORCES. 165 

the strings oa, ob, oc, od, oe. No further modification of the 
process is necessary. 

193. Moment of Inertia Determined from Area of Funicular 
Polygon. — In Fig. 65, the moments of the given forces are 
represented by the intercepts A' B\ B'C\ CD', D' E' , each 
multiplied by the pole distance H. The moment of inertia of 
any force, as AB, is equal to the moment of a force represented 
by the corresponding intercept as A' B' , supposed to act in the 
line ab. Now, by definition (Art. 175) the moment about the 
axis QR of a force equal to A' B' acting in the line ab is equal 
to double the area of the triangle A' i B' ; hence the moment of 
the force HxA'B' is equal to double the area of that triangle 
multiplied by//. Similarly, the moment of a force HxB'C, 
acting in the line be, is equal to double the area of the triangle 
B' 2 C multiplied by //. Applying the same reasoning to each 
force, we see that the sum of the moments of the assumed 
forces {HxA'B', NxB'C, etc.) is equal to 2 //times the sum 
of the areas of the triangles A^ i B', B'zC, C 3D', D' 4E'. 
In adding these triangles, each must be taken with its proper 
sign, corresponding to the sign of the moment represented by 
it. Thus, the moments of A'B', B'C, and D'E' all have the 
same sign, while the moment of CD' has the opposite sign. 
Hence we must have for the sum of the moments, 

area A' i /?' + area B' 2 C —SLve^L C 3 /^' + area D' 4E', 

which is equal to the area of the polygon A' i 2 34E'. Hence 
this area, multiplied by 2 //, gives the moment of inertia of the 
required system of forces. 

It may sometimes be convenient to apply this principle in 
determining moments of inertia, the area being determined by 
use of a planimeter, or by any other convenient method. It 
should be noticed that if H is taken equal to the sum of the 
given forces (AE), twice the area of the funicular polygon is 
equal to the square of the radius of gyration. If //= l AE, the 



1 66 



GRAPHIC STATICS. 



square of the radius of gyration is equal to the area of the 
polygon. 

194. Determination of Product of Inertia of Parallel Forces. 

— Assume the points of application of the forces to be in the 
plane containing the two axes. If the moment of any force 
with respect to one axis be found, and a force equal in magni- 
tude to this moment be assumed to act at the point of applica- 
tion of the original force, then the moment of this new force 
with respect to the second axis is equal to the product of 
inertia of the given force for the two axes. 

Thus, let ab, be, cd, de (Fig. 66) be the points of application 
of four parallel forces, and let their product of inertia with 
respect to the axes QR, 57" be required. Draw ABCDE, the 




force polygon for the given forces, assumed to act parallel to 
QR. Choose a pole O, the pole distance being preferably 
taken equal to AE, or some simple multiple of AE, and draw 
the funicular polygon as shown, prolonging the strings to inter- 
sect QR in the points A\ B\ C\ D\ E\ Now assume a series 



MOMENTS OF INERTIA OF FORCES. 



167 



of forces equal to A'B', B'C\ etc., each multiplied by //", to act 
at the points ad, be, etc., and determine their moments with 
respect to the axis ST. To find these moments, draw lines 
through ab, be, etc., parallel to ST, and draw a funicular poly- 
gon for the assumed forces taken to act in these lines. The 
force polygon is obtained by revolving the line A'B^C'D^E' until 
parallel with ST, and is the line A\ B\ C\ D\ E\ in the figure. 
The strings o^a\ <?V of the second funicular polygon intersect 
ST in the points A\ and j£"". Hence, calling //' the second 
pole distance, the product of inertia of the given system is 
equalto/^xi7'x^"£". 

195. Product-Radius. — If //'be taken equal to AE (Fig. 6()), 
/f'x^"ir" is equal to the square of the product-radius (Art. 
189). Hence the product-radius can be found by a construc- 
tion exactly like that employed in finding the radius of gyra- 
tion. Thus (Fig. ^6) take LM=H' and MN=A''E"; make 
LN the diameter of a semicircle, and draw from M a line per- 
pendicular to LN, intersecting the semicircle in P. Then 
'mP'=LM X2MN= H'xA''E^'\ hence MP=e, the product- 
radius. 

196. Relation between Moments of Inertia for Parallel Axes. 

— Proposition. — The moment of inertia of a system of par- 
allel forces with reference to any axis is equal to its moment of 
inertia with respect to a parallel axis through the centroid of 
the system plus the moment of inertia with respect to the given 
axis of the resultant applied at the centroid of the system. 

Let Pi, Po, etc., represent the forces ; X\, Xo, etc., the dis- 
tances of their points of application from the central axis ; and 
a the distance of this central axis from the given axis. Calling 
the required moment of inertia A, and the moment of inertia 
with respect to the axis through the center of gravity A', we 
have 

A=P, {a-^XiY-^PoAa^-x,)-^-" 

= aAP,-{-P,+ "')^-2a{P^r,-\-P,x,+ ---)+P,x,' + P,xi+-- 



l68 GRAPHIC STATICS. 

Now PyXy^P.2X\_^ is the algebraic sum of the moments of 

the given forces with respect to the axis through their centroid, 
and is equal to zero ; and P^Xx^P^x^"'^ = A[. Hence 

which proves the proposition. 

Radii of gyration. — Let k = radius of gyration of the sys- 
tem with respect to the given axis, and k^ the radius of gyra- 
tion with respect to the central axis, and we have 

A = {P,^P -V-')k\ 

A'={P,-\-P,^...)k'\ 

Hence the equation above deduced may be reduced to the form 

k' = k''' + a\ 

197. Products of Inertia with Respect to Parallel Axes. — 

Propositioji. — The product of inertia of a system of parallel 
forces with reference to any two axes is equal to the product 
of inertia with reference to a pair of central axes parallel to the 
given axes, plus the product of inertia of the resultant (acting 
at the centroid) with reference to the given axes. 

Let /*!, Poy etc., be the magnitudes of the given forces ; {px, ^1), 
(/2j <■]'>), etc., the distances of their points of application from 
the central axes parallel to the two given axes ; {a, b) the dis- 
tances of the centroid of the system from the given axes. Let 
A and A' be the products of inertia of the system with respect 
to the given axes and the parallel central axes respectively. 

Then 

A' = P,p,q, + P,p.,q,-{--" 

A =Pi{p, + a){qi + b) +/'2 (/2 + «)fe + ^) + ••• 
= (P^P^qi + P2p2q2 +"')+a {P,q, + P.q. + • • •) 

But P\qi-\-P2q-2^ — =0; and P\px-\-P^P^A — =0; since each of 
these expressions is the sum of the moments of the given forces 



MOMENTS OF INERTIA OF PLANE AREAS 169 

with respect to an axis through the centroid of the system. 
Hence, 

A=A'-\- {P,JrP,+ '-')ab, 

which proves the proposition. 

liA = {P, + P,+ --) r and A^={P, + P,+ ...)c'\ we have 

c' = c'~-\-ab. 

From the above proposition it follows that if the axes have 
such directions that the product of inertia with reference to the 
central axes is zero, the product of inertia with reference to the 
given axes is the same as if the forces all acted at the centroid. 
When this condition is known to be satisfied, then for the pur- 
pose of finding the product of inertia the system of forces may 
be replaced by their resultant. 

It follows also, in the case when the product of inertia for the 
central axes is zero, that if one of the given axes coincides with 
the parallel central axis, the product of inertia for the given 
axes is zero ; for in this case either a or b i?, zero, and hence 
ab{Pi + P2-{ — ) is zero. Therefore, 

If the product of inertia of a system is zero for two axes, A' 
and A" , one of which (as A') contains the centroid of a system, 
then the product of inertia is also zero for A' and any axis 
parallel to A". 

§ 2. Moine7its of Inertia of Plane Areas. 

198. Elementary Areas Treated as Forces. — If any area be 
divided into small elements, and a force be applied at the 
centroid of each element numerically equal to its area, the 
moment of inertia of this system of forces will be approximately 
equal to that of the given area. The approximation will be 
closer the smaller the elementary areas are taken. If the ele- 
ments be made smaller and smaller, so that the area of each 
approaches zero as a limit, the moment ,of inertia of the sup- 
posed system of forces approaches as a limit the true value of 
the moment of inertia of the given area. 



170 



GRAPHIC STATICS. 



It is seen, then, that most of the general principles which 
have been stated regarding moments of inertia of systems of 
forces are equally applicable to moments of inertia of areas. 
The practical application of these principles, however, and 
especially the graphic constructions based upon them, are less 
simple in the case of areas than of systems of forces such as 
those already treated. The reason for this is that the system 
of forces which may be conceived to replace the elements of 
area consists of an infinite number of infinitely small forces, 
with which the graphic methods thus far discussed cannot 
readily deal. Problems of this class are most easily treated by 
means of the integral calculus, especially when the areas dealt 
with are in the form of geometrical figures. It is possible, 
however, by graphic methods to determine approximately the 
moment of inertia of any plane area ; and in many cases exact 
graphic solutions of such problems are not difficult. The proof 
of these methods is often most easily affected algebraically. 

199. Moments of Inertia of Geometrical Figures. — The appli- 
cation of the integral calculus to the determination of moments 
of inertia will not be here treated. But the values of the 
moments of inertia of some of the common geometrical figures 
are of such frequent use that the more important of them will 
be given for future reference. The moment of inertia is in each 
case taken with respect to a central axis, and will be repre- 
sented by /, while the radius of gyration will be called k. 

Rectangle. — Let b and d be the sides, the axis being parallel 
to the side b. Then 

12 ' 12 

Triangle. — Let b and d be the base and altitude. Then for 
an axis parallel to the base, 

/=M! . k'^ = — 
36 ' i8' 



MOMENTS OF INERTIA OF PLANE AREAS. 171 

For an axis through the vertex, bisecting the base, k'^ = — , 

where b^ is the projection of the base on a hne perpendicular to 
the axis. 

Citxle. — Let d be the diameter. Then 

64 ' 16 

For a central axis perpendicular to the plane of the circle^ 

Ellipse. — Let a and b be the semi-axes. Then for an axis 
parallel to a, 

j_iTab^ ro_b'^ 
I — , /v- — — 

4 4 

For an axis parallel to b, 

j_ ird^b . A2 _ <^^ 
4 ' 4 

For a central axis perpendicular to the plane of the ellipse, 

4 ' 4 

Graphic constricction for radius of gyration. — Whenever k'^ 
can be expressed as the product of two known factors, the value 

of k can be found by the construction already used in Art. 191. 

^2 
Thus, in case of a rectangle, for which 1^^ = — , we may put 

j^i^d d ^j^^^ .^ .^ p. ^^ ^^^ ^^^^ LM=-, MN=-, the 

3 4^ 4 3 

construction there shown will giv^e MP as the value of k. For 

the triangle, the axis being parallel to the base, we have 

B = - • -, and the same construction is applicable. For the 
^ 6 

. ^' ^' 

axis through the vertex bisectin£c the base, 1^^=— • — • 
^ "" 4 6 



1/2 



GRAPHIC STATICS. 



200. Product of Inertia. — General principles. — Products of 
inertia of areas are determined by means of the integral calculus 
in a manner similar to that employed for moments of inertia. 
The following fundamental principles regarding products of 
inertia of geometrical figures will be found useful : 

(i) With reference to two rectangular axes, one of which is 
an axis of symmetry (Art. 179), the product of inertia is zero. 
For it is manifest that the products of inertia of two equal 
elements, symmetrically placed with reference to one of the 
axes, are numerically equal but of opposite sign. Hence, if 
the whole area can be made up of such pairs of elements, the 
total product of inertia is zero. 

(2) If the two axes are not rectangular, but the area can be 
divided into elements such that for every element whose dis- 
tances from the axes are /, ^, there is an equal element whose 
distances are /, —q, or — /, q, the product of inertia is zero. 
This includes the preceding as a special case. 

201. Products of Inertia of Geometrical Figures. — In each of 
the following cases the product of inertia is zero : 

A triangle, one axis containing the vertex and the middle 
point of the base, the other being any line parallel to the base. 

A parallelogram, the axes being parallel to the sides and one 
axis being central. This includes the rectangle as a special 
case. 

An ellipse, the axes being parallel to a pair of conjugate 
diameters, and one axis being central. This includes, as a 
special case, that in which one axis is a principal diameter and 
the other is any line perpendicular to it ; and under this case 
falls also the circle. 

202. Approximate Method for Finding Moment of Inertia of 
Any Area. — To apply the method of Art. 190 to the determina- 
tion of the moment of inertia of a plane area, we should strictly 
need to replace the area by an infinite number of parallel forces, 
proportional to the infinitesimal elements of the given area, and 



MOMENTS OF INERTIA OF PLANE AREAS. 



7j 



with points *of application in these elements. If, instead, we 
divide the given figure into finite portions whose several areas 
are known, and assume forces proportional to those areas to 
act at their centroids, we may get an approximate value for the 
moment of inertia, which will be more nearly correct the smaller 
the elements. This will be illustrated by the area shown in 
Fig. ^J. 

Let QR be the axis with reference to which the moment of 
inertia is to be found, in this case taken as a central axis. 




Fig. 67" 



Divide the figure into four rectangular areas as shown, and 
assume forces numerically equal to these areas to act at their 
centroids parallel to QR. The force polygon is ABCDE. 
Draw the funicular polygon corresponding to a pole (9, and let 
the successive strings intersect QR in A\ B\ C , D\ E'. Take 
this as a new force polygon, and with any convenient pole 
distance draw a second funicular polygon, using the same lines 
of action. Let the first and last strings intersect QR in A" 
and £■". Then A" E" multiplied by the product of the two 
pole distances gives the moment of inertia of the four assumed 
forces, and approximately the moment of inertia of the given 



174 



GRAPHIC STATICS. 



figure. If the first pole distance is taken equal to AE (as is 
the case in Fig. 6y), the radius of gyration may be found by the 
construction of Art. 191. Thus in Fig. Gy, MP is the radius 
of gyration as determined by this method. 

A more accurate result may be reached by dividing the area 
into narrower strips by lines parallel to QR ; since the narrower 
such a strip is, the more nearly will the distance of each small 
element from the axis coincide with that of the centroid of the 
strip. If the partial areas are taken as narrow strips of equal 
width, the forces may be taken proportional to the average 
lengths of the several strips. 

203. Accurate Methods. — If the given figure can be divided 
into parts, such that the area of each is known, and also its 
radius of gyration with respect to its central axis parallel to the 
given axis, the above method may be so modified as to give an 
accurate result. Two methods will be noticed. 

(i) When the axis is kitown at the stai't. — Let the line of 
action of the force representing any partial area be taken at a 
distance from the given axis equal to the radius of gyration of 
that area with reference to the axis. If this is done, it is 
evident that the moment of inertia of the system of forces is 
identical with that of the given area. When the axis is known, 
the position of the line of action for any force may be found as 

follows : Let QR (Fig. 6Z) be the given axis, 
and Q R^ a parallel axis through the centroid 
of any partial area. Draw KL perpendicu- 
lar to QR, and lay off LM equal to the 
radius of gyration of the partial area with 
respect to Q R^ . Then KM is the length 
of the radius of gyration with respect to QR. (Art. 196.) Take 
KN^^M'A to KM and draw Q^R'^ through N parallel to Q R' ; 
then Q^'R" is to be taken as the line of action of the force 
representing the partial area in question. 

(2) When the axis is at first tinhtown. — The method to be 




MOMENTS OF INERTIA OF PLANE AREAS. 



1/5 



employed in this case is to let the force representing any partial 
area act in a line through the centroid of that area ; and then 
assume the force representing its moment to act in such a line 
that the moment of this second force shall be numerically 
equal to the moment of inertia of the partial area. This line 
may be found as follows : Let k represent the radius of gyra- 
tion of the partial area with respect to its central axis parallel 
to the given axis, and a the distance between the two axes. 
Then the moment of inertia of the partial area with respect 
to the given axis is A {a'^-\-k'^), if A represents the area. But 

A {a^-\-k^)=Aala-\ — ). Hence, if a force numerically equal to 

A is assumed to act with an arm a, then a force equal to its 

P . 
moment Aa must act with an arm a-\ — in order that its moment 

a 
may equal A {a^-\-P). 

The distance a-\ — can be found by a simple construction. 
a 

Let QR and Q R^ (Fig- 69) be the given axis and the central 
axis respectively. Draw KL 
perpendicular to QR and lay off 
LM equal to k. From M draw 
a line perpendicular to KM^ in- 
tersecting KL produced at N. 

Then KN=a+-- For, in the 
a 




Fig. G9 



similar triangles KNM, KML, we have 
KN KM 



KM KL 



or KN= 



Km 

KL ' 



But KL=a, and KM^ = a^ + k'^; hence 

a a 

This second method is more useful than the first, because in 
applying it the first funicular polygon can be drawn before the 
position of the inertia-axis is known. Thus, a very common 



1/6 



GRAPHIC STATICS. 



case is that in which the moment of inertia of an area is to be 
found for a central axis, whose direction is known, while at the 
outset its position is unknown because the centroid of the area 
is unknown. If the second method be employed, the first 
funicular polygon can be drawn at once, and serves to locate 
the required central axis, as well as to determine the moments 
of the first set of forces as soon as the axis is known. The 
central axis and the moment of inertia with respect to it are 
thus determined by a single construction. 

Example. — The method last described is illustrated in Fig. 
70. The area shown consists of two rectangles, the centroids 
of which are marked ab and be. The moment of inertia is to 
be found for a central axis parallel to the longer side of the 
rectangle ab. 




Fis. "TO 



We draw through ab and be lines parallel to the assumed 
direction of the axis, and take these for the lines of action of 
forces AB, BC, proportional to the areas of the two rectangles. 
ABC is the force polygon, and the pole distance is taken equal 



MOMENTS OF INERTIA OF PLANE AREAS. 



177 



to AC. The intersection of the strings oa, oc, determines a 
point of the required central axis ac. The moments of the 
given forces are proportional to A^B\ B'C. We now have to 
find the lines of action for the forces A'B\ B'C, in accordance 
with the method above described. 

Take any line perpendicular to the axis ac, as VQ, the side 
of one of the rectangles. From R, the point in which this line 
intersects the vertical line through be, lay oK RT equal to the 
central radius of gyration of the rectangle whose centroid is 
be. To find this central radius of gyration, we know that its 

value is V — (Art. 199), where d is the length of the side of 

the rectangle perpendicular to the axis. Hence we take 

QR = —,RS=—, and make QS the diameter of a semicircle, 
3 4 

intersecting the vertical line be \n T\ then RT is the required 
radius of gyration of the rectangle with respect to a central 
axis. Now draw from T a. line perpendicular to VT, intersect- 
ing VQ in 17; then the line b'e' drawn through 6^ parallel to the 
given axis is the line of action of the force B'C. 

By a similar construction applied to the other rectangle, a'b' 
is located as the line of action of the force A'B'. The second 
funicular polygon is now drawn, and the points A", C" are 
found by the intersection of the strings o'a', o'e' with the axis. 
Hence the moment of inertia of the given area is equal to 
A"C" xHxN'. In the figure H is made equal to AC, hence 
the radius of gyration can be determined by the usual con- 
struction, and its length is found to be MP. 

204. Moment of Inertia of Area Determined from Area of 
Funicular Polygon. — The method given in Art. 193 for finding 
the moment of inertia of a system of forces by means of the 
area of the funicular polygon may be applied with approximate 
results to the case of a plane figure. If the forces are taken 
as acting at the centroids of the areas they represent, then to 



178 



GRAPHIC STATICS. 



get good results these partial areas should be taken as narrow 
strips between lines parallel to the axis. 

If the lines of action are determined as in the first method 
of the preceding article, then the area enclosed by the funic- 
ular polygon and the axis represents accurately the required 
moment of inertia. 

205. Product of Inertia Determined Graphically. — To deter- 
mine the product of inertia of any area, let it be divided into 
small known parts, and let parallel forces numerically equal to 
the partial areas be assumed to act at the centroids of these 
parts. The product of inertia of these forces may then be 
found as in Art. 194, and its value will represent approximately 
the product of inertia of the given area. 

If the partial areas can be so taken that the product of 
inertia of each with reference to axes through its centroid par- 
allel to the given axes is zero, the method here given is exact. 
(Art. 197.) 

If the partial areas are taken as narrow strips parallel to one 
of the axes, the condition just mentioned will be nearly fulfilled ; 
for the product of inertia of each strip for a pair of axes 
through its centroid, one of which is parallel to its length, will 
be very small. 



CHAPTER X. CURVES OF INERTIA. 

§ I. General Principles. 

206. Relation between Moments of Inertia for Different Axes 
through the Same Point. — The moment of inertia with respect 
to any axis through a given point can be expressed in terms of 
the moments and product of inertia for any two axes through 
that point. It is necessary here to use algebraic methods, but 
the results reached form the basis of graphic constructions. 

Let OX, 6^F(Fig. 71) be the two given axes; 0, the angle 
included between them ; P], Pi, etc., the forces of the system ; x^ , 
J'', the coordinates of the point 
of application of any force P, 
referred to the axes OX, OY; 
/, {/, the perpendicular distances 
of the same point from O V and 
OX respectively, so that 

/=.r'sin^, q=y sm6. 

Let a', b\ and c' be quantities defined by the equations 

A;r/2 + P2^2''+ '"_tPx''' 




a —■ 



P ^P,^ 



^P 



^P 

,0 tPx'y 

C '= — ^ —^ — • 



179 



l8o GRAPHIC STATICS. 

Then 

a'"^ sm^^ = = — ^ 

XP 1.P 

^^,^,^,0^tpyHin^e^iPl^ 

tP XP 

it,2 ^ _ ^^^y sin^ ^ XPpq . 
%P %P ' 



c''^ SI 



and a' sin 6, b' sin 6, c' sin are respectively the radius of 
gyration with respect to O V, the radius of gyration with respect 
to OX, and the product-radius (Art. 189) with respect to OX 
and OV. 

The moment of inertia (/) and radius of gyration (k) of the 
system for the axis OM, making an angle </> with OX, may now 
be computed as follows : 

Let s = perpendicular distance of the point of application of 

any force P from OM. Then from the geometry of the figure 

it is seen that 

s =y sin (6 — (f)) —x^ sin ^. 
Hence 

/= tPs^=tPy^ sin2 (61 -(/))- 2 tPx'y' sin (<9 - 0) sin <f> 

-{-tPx''^ s'm^ cf) ; 
or, 

I=b'^tP . sin2 {e-(f>)-2 c^'^tP . sin {6-<i>) sin <j> 

+^'2^P.sin2(^, 

the factors involving 6 and <^ being constant for all terms of 
the summation. Hence 

y^2=_^ = <^'2sin2((9-c/))-2^^2sin((9-</))sin<^ + ^'2sin2(^ . . (i) 

From these equations / and k may be computed if a', b\ and 
c^ are known ; that is, if the moments and product of inertia for 
the two axes (9Xand 6^ Fare known. 

Special case. — If 6 = go°, the equation (i) becomes 

k^ = b'"^ cos^ (^ — 2 c''^ ?,\n (f) cos (j> -\- a'"^ s\n^ (j) (2) 



GENERAL PRINCIPLES. l8i 

207. Products of Inertia for Different Axes through the Ori- 
gin. — The product of inertia with respect to OM and OX may 
be found as follows : 

Let A = the required product of inertia ; then 

A=:tPqs=%Py sin l9 [/ sin((9-(^)-y sin 0] 
= 2P [/^ sin e sin (^-(^) -Jir'/ sin 6 sin 0] 
= [^'2 sin 6 sin (6' - <^) - c'^ sin (9 sin </>] 5!^. 
Let /^ = product-radius for axes OM a.nd OX; then 

/22 = ^ = ^'2 sin (9 sin {6 - (f>) - c'''- sin (9 sin <^. 

Special case. — The axis OM may be so chosen that A—o. 
This will be the case if 

^'2 sin ((9 -(/))= ^'2 sin0. 

208. Inertia Curve. — If on OM (Fig. 71) a point J/ be taken 
such that the length 6^ J/ depends in some given way upon the 
value of k, and if similar points be located for all possible direc- 
tions of OM^ the locus of such points will be a curve which is 
called a curve of inertia of the system for the center O. 

The form of the curve will depend upon the assumed law 
connecting (9Jfwith k. 

209. Ellipse or Hyperbola of Inertia. — The simplest curve is 

obtained by assuming OM to be inversely proportional to k. 

d^ sin^ Q 
Let OM—r, and take r'^= — — ^ — > where d"^ is a positive 

ft 

quantity, so that d always represents a real length, posirive or 
negative. 

Equation (i) of Art. 206 then becomes 
d^ sin2 (9 = ^'2;^ sin2 ((9 -</>)- 2 ^'2^2 sin {6-^) sin (/) + ^'2;.2 sin2 0, 

which is the polar equation of the inertia-curve, r and <^ being 
the variable coordinates. Let x, y be coordinates of the point 
M referred to the axes OX, O V. Then 

r _ X _ y 
sin ^ sin(^ — (^) sin </> 



l82 GRAPHIC STATICS, 
whence ^ ; ~^' =x'^, 

r^ sin^ (j) _ 2 
sin2 6> ~^' 

r^ sin (6 — ^) sin (^ _ 

and the equation becomes 

^^ = l^''ij^2_2c'^:^yj^a'y- (3) 

The form of this equation shows that it represents a conic 
section whose center is at the origin of coordinates O. This 
conic may be either an ellipse or a hyperbola. 

The equation will be discussed more fully in a later article. 
One fact may, however, be here noticed. 

If the moment of inertia / is positive for all positions of the 
axis, the radius of gyration k will be real, whatever the value 
of (j). But r, the radius vector of the curve, will be real when k 
is real ; hence in this case the curve is an ellipse. This is 
always the case if the given forces have all the same sign. 

If the forces have not all the same sign, it is possible that 
the value of / may have different signs for different directions 
of the axis. If this is so, certain values of (/> make k (and 
therefore r) imaginary. In this case the curve is a hyperbola. 

The most important case is that in which the given forces 
have all the same sign which may be taken as plus, so that the 
moment of inertia is always positive, and the curve an ellipse ; 
and to this case the discussion will be confined. 

§ 2. Inertia-Ellipses for Systems of Forces. 

210. Properties of the Ellipse. — In discussing ellipses of in- 
ertia use will be made of certain general properties of the 
ellipse, which, for convenience of reference, will be here sum- 
marized. For the proof of the propositions stated the reader 
is referred to works on the conic sections. 



INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. 183 

(i) The equation 

represents an ellipse if B^—AC is negative; a hyperbola if 
B^ — AC is positive. (Salmon's Co7ttc Sections, p. 140.) The 
coordinate axes may be either rectangular or oblique. 

(2) Two diameters of an ellipse are said to be conjugate to 
each other if each bisects all chords parallel to the other. If 
the axes of coordinates coincide with a pair of conjugate 
diameters, the lengths of which are 2 a' and 2 b', the equation 
of the curve is 

A particular case of this equation is that in which the coordi- 
nate axes are rectangular, being the principal axes of the curve ; 
in which case we may write a and b instead of a' and b^. 

(3) In an ellipse, the product of any semi-diameter and the 
perpendicular from the center on the tangent parallel to that 
semi-diameter is constant and equal to ab. That is, if r is any 
radius vector of the curve drawn from the center, and / the 
length of the perpendicular from the center to the parallel 
tangent, we have 

pr=ab 

where a and b are the principal semi-axes of the curve. 

(4) Let a' and b' be conjugate semi-diameters. Then each is 
parallel to the tangent at the extremity of the other. Hence 
the length of the perpendicular from the center to the tangent 
parallel to a' is b^ sin 0, where 6 is the angle included between 
a' and b'. Therefore from the preceding paragraph, 

a'b' sin d = ab. 

(5) An elHpse can be constructed, when a pair of conjugate 
diameters is known, as follows : 





/ ' 




B C 


K ^ 




1-' /'- /. 


/ y / 




Fi 


B 


72 



184 GRAPHIC STATICS. 

Let AA', BB' (Fig. 72), be the conjugate diameters, O being 
the center of the ellipse. Complete the parallelogram OBCA. 

Divide OA and CA into parts 
proportional to each other, be- 
ginning at O and C. Through 
the points of division of OA 
draw lines radiating from B' y 
and through the points of divi- 
sion of CA draw lines radiating 
from B. The points of inter- 
section of the corresponding lines in the two sets are points of 
the ellipse. In a similar way, the other three quadrants may 
be drawn. (The location of one point is shown in the figure.) 

A convenient way to locate the corresponding points of 
division on OA and CA is to cut these lines by lines parallel to 
the diagonal OC 

211. Discussion of Equation of Inertia-Curve. — We will now 
examine the equation of the inertia-curve, 

with reference to the properties of the ellipse above enumerated. 

(i) If a''^b''^ — c"^ is positive, the equation denotes an ellipse. 
This cannot be the case if a'"^ and b''^ have opposite signs. But 
from the definitions of a''^ and b''^ (Art. 206) it is seen that their 
signs are the same as those of the moments of inertia for Y and 
X axes respectively. Hence, if there are any two axes through 
the assumed center for which the moments of inertia have 
opposite signs, the inertia-curve is a hyperbola. 

If the moment of inertia has the same sign for all axes 
through the assumed center, the curve is an ellipse. For, since 
c''^= ^ (Art. 206), c^ may be made zero by choosing the 

axes so that the product of inertia with respect to them is 
zero ; and if c' is zero, and a''^ and b^'^ have the same sign, the 
quantity a^^b'- — c'^ is positive. 



INERTIA-ELLIPSES FOR SYSTEMS OF FORCES. 185 

This agrees with the conclusion stated in Art. 209. 
We shall here deal only with ellipses of inertia. 

(2) If c' = o, the coordinate axes are conjugate axes of the 
curve. But the condition c' = o means that the product of 
inertia for the two axes is zero. Hence any two axes for which 
the product of inertia is zero are conjugate axes of the inertia- 
curve. (This is true whether the curve is an ellipse or a 
hyperbola.) 

(3) By the law of formation of the inertia-conic (Art. 209), 
the length of the radius vector lying in any line is inversely 
proportional to the radius of gyration with respect to that line. 
But by (3) of the last article, the perpendicular from the center 
on the tangent parallel to any radius vector is inversely propor- 
tional to the length of that radius vector. Hence the perpen- 
dicular distance between any diameter and the parallel tangent 
is directly proportional to the radius of gyration with respect to 
that diameter. The curve may be so constructed that the 
length of this perpendicular is equal to the radius of gyration, 
as follows : 

From Art. 209, we have 

j_d^ sin 6 
r 

and from (3) and (4) of the last article we have 

, ab a^b^ sin 

P = ^ = ' 

r r 

if a! and b^ are conjugate semi-diameters. Now take 

^^ sin 6 = ab = a'b^ sin 6y 

or d' = a'b', 

and we have k=p, 

and the equation of the curve becomes (since c^ = o when the 
axes are conjugate) 

b'x--\-a y =a . 



i86 



GRAPHIC STATICS. 



If the equation be written in this form, a' and b' having the 
meanings assigned in Art. 206, the radius of gyration about any 
axis through the center of the ellipse is equal to the perpendicular 
distance betweeji the axis and the parallel tangent to the ellipse. 

Hereafter we shall mean by inertia-ellipse the curve obtained 
by taking d^ = a'b^ as above described, so that the radius of gyra- 
tion for any axis can be found by direct measurement when a 
parallel tangent to the ellipse is known. 

212. To Determine Tangents to the Inertia-Ellipse for Any- 
Center. — Let the radius of gyration {k) be found for any 
assumed axis through the given center by one of the methods 
already described (Arts. 202 and 203). Then two lines par- 
allel to the axis and distant k from it, on opposite sides, will be 
tangents to the inertia-ellipse. 

213. To Construct the Inertia-Ellipse, a Pair of Conjugate 
Axes Being Known in Position. — If the positions of two axes 
conjugate to each other can be found, the ellipse can be drawn 
by the following method : 

Determine the radius of gyration for each of the two axes 
and draw the corresponding tangents as in the preceding 
article ; then proceed as follows : 

Let XX', YV (Fig. 73) be the given axes, and let the four 
tangents determined as above form the parallelogram PQRS. 

Let A, A', B, B' be the points 
in which these tangents in- 
tersect the axes XX', YY'. 
Then, since each diameter is 
parallel to the tangents at the 
extremities of the conjugate 
Y' Fig. 73 diameter. A, A', B, B' are the 

extremities of the diameters lying in the given axes. The 
elhpse can now be constructed as explained in Art. 210 
(Fig. 72). 

This method of constructing the inertia-ellipse is useful 




INERTIA-CURVES FOR PLANE AREAS. 187 

whenever the given system has a pair of conjugate axes which 
can be located by inspection. 

214. Central Ellipse. — It is evident that an inertia-curve 
can be found with its center at any assumed point. That 
ellipse whose center is the centroid of the given system is called 
the central ellipse for the system. 

Since the central ellipse gives at once the radius of gyration 
for every axis through the centroid of the system, it enables us 
to determine readily the radius of gyration for any axis what- 
ever, by means of the known relation between radii of gyratioa 
for parallel axes. (Art. 196.) 

§ 3. Iiiertia-Cnrves for Plane Areas. 

215. General Principles. — The principles deduced in the 
treatment of inertia-curves for systems of forces are all true for 
the case of plane areas. But special difficulties arise in dealing 
with areas, because of the fact that the system of forces equiva- 
lent to any area consists of an infinite number of forces. The 
principles already developed are, however, sufficient to deal at 
least approximately with all areas, and accurately with many 
cases. 

216. Inertia-Curve an Ellipse. — Since the forces conceived 
to replace the elements of area (Art. 198) have all the same 
sign, the value of k'^ is always positive, and the inertia-curve is 
always an ellipse. (Arts. 209 and 211.) 

217. Cases Admitting Simple Treatment. — Whenever a pair 
of conjugate diameters can be located, and the radius of gyra- 
tion determined for each, the inertia-ellipse can be at once 
drawn as in Art. 213. This will be the case whenever it is 
possible to locate readily a pair of axes for which the product of 
inertia is zero. 

(i) If there is an axis of symmetry, this and any line perpen- 
dicular to it are a pair of conjugate axes (and in fact the principal 
axes) of the inertia-ellipse whose center is at their intersection. 
(Art. 200.) 



1 88 GRAPHIC STATICS. 

(2) If two axes can be located in such a way that for every 
element of area whose distances from the axes are/, q, there is 
an equal element whose distances are p, —q, or — /, q, the 
product of inertia is zero for the two axes, and these are there- 
fore a pair of conjugate axes of the inertia-ellipse whose center 
is at their intersection. This of course includes the case when 
there is an axis of symmetry. (Art. 200.) 

When a pair of conjugate axes is known, the radius of gyration 
is to be found by one of the methods of Art. 202 or Art. 203 ; 
the ellipse can then be drawn exactly as explained in Art. 213. 

If a pair of conjugate axes cannot be located by inspection, 
the inertia-ellipse cannot be so readily constructed. Such cases 
will not be here treated. 

As examples of areas, in which the principal axes of the 
inertia-curve can be located by inspection, may be mentioned 
the cross-section of the I-beam, the deck-beam, the channel-bar, 
and other shapes of structural iron and steel. 

Many geometrical figures possess axes of symmetry. In 
others a pair of conjugate axes can be located by principle (2). 
Some of these will be discussed in the next article. 

Example. — Draw the central ellipse for the deck-beam sec- 
tion shown in Fig. 74. 



-1-%T»' 



as 



0~^~ [Suggestions. — Since there is an axis of symmetry, this 

contains one of the principal axes of the eUipse. The other 
can be drawn as soon as the centroid of the section is 
known. Find the radius of gyration for each axis by 
Art. 202, and then construct the elUpse as explained in 
"^2- Art. 213.] 



2 1 8. Central Ellipses for Geometrical Fig- 
ures. — In many of the simple geometrical 
figures, not only can a pair of conjugate 
axes be located by inspection, but the radius 
of gyration for each of these axes can be found by a simple 
construction, so that the central ellipse can be readily drawn. 
Some of these cases will be here summarized. 




INERTIA-CURVES FOR PLANE AREAS. 



189 




(i) Parallelogram. — Let ABCD (Fig. 75) be the parallelo- 
gram; then XX\ YV, drawn through the centroid parallel to 
the sides, are a pair of 
conjugate axes of the cen- ^ 

tral ellipse. Let AB = b, 
BC=dj and let h= the 
perpendicular distance be- 
tween AB and DC. The 
moment of inertia of the 
parallelogram with re- 
spect to the axis XX' is 
equal to the moment of inertia of a rectangle of sides b and //. 
Hence k^, the square of the radius of gyration for this axis, is 

— ■• The length of k can be found by the construction used in 

case of the rectangles in Fig. 70. The following modification 
of the method is, however, more convenient : 

Make EF=\BC, EG = \ BC, and draw a semicircle with FG 
as a diameter. From E draw a line perpendicular to BCy inter- 
secting the semicircle at /. Lay off EH=EI\ then a line 
through H parallel to XX^ is a tangent to the central ellipse. 
For by construction, 

BC 



EH=EI-- 



V12 



And since the projection of BC on a line perpendicular to XX\s> 
equal to //, the projection of EH on the same line is equal to 



V 



12 



that is to k. 

The tangent parallel to the side BC may be found in a simi- 
lar way. It may, however, be located more simply as follows : 
It is evident that the distance between YY' and a tangent par- 
allel to it, measured along AB, bears the same ratio to AB that 
EH does to BC Hence, the parallelogram formed by the four 
tangents, two parallel to XX' and two parallel to YY\ is simi- 
lar to the parallelogram ABCD. 



190 



GRAPHIC STATICS. 



Fig.' 75 shows this parallelogram and also the ellipse. 

(2) Triangle. — Let ABC (Fig. j6) be the triangle ; b = 
Y length of the base BC\ b' = 

length of projection of BC on a 
line perpendicular to AD ; d = a.\- 
titude measured perpendicular to 
BC. AD and a line through the 
centroid parallel to BC are a pair 
of conjugate axes of the central 
ellipse (Art. 217). From Art. 
199, the radius of gyration for a 
central axis parallel to BC is 

Let XX' be this axis, and H the point in which 
3 6 
it intersects ^C Then HC=\AC. Take HK=\AC=\HC, 

and make KC the diameter of a semicircle. From H draw HI 

perpendicular to AC, intersecting the semicircle at /. Make 

HL — HI\ then the line through L parallel to XX^ is a tangent 

to the central ellipse. For the radius of gyration with respect 

to XX is to HL as the altitude d is to AC. 

Again, for the axis AD, the radius of gyration is A/ — (Ai 

199). Make 

diameter of a semicircle. 




.rt. 



DE=\ BC and DF-- 



24 



\ BC, and take EF as a 
From D draw a line perpendicular to 
BC, intersecting the semicircle in M\ and make DG = DM; then 
a line from G parallel to AD is a tangent to the central ellipse. 
The figure shows the parallelogram formed by the two tan- 
gents parallel to XX' and the two parallel to AD, and also the 
central ellipse. 



(3) Ellipse. — From Art. 199, the radii of gyration of an 
ellipse with respect to the two principal diameters are -| a 
and J b. Hence the central ellipse of inertia is similar to the 
given ellipse, its semi-axes being J a and ^ b. A special case 
of this is a circle, for which the central curve is a circle whose 
radius is half that of the given circle. 



INERTIA-CURVES FOR PLANE AREAS. 



191 



Fis. TT 







Y 




^ 


/— 


C 


^ 


X' / 


(F 




\ X 


1 


A 


) 


\ 


I 


,/v^!\^ 


G_^ 


\ 


1 






1 


A E 


D 


B 






Y 





(4) Semicircle. — Let ABC (Fig. jj) be the semicircle, O 
being the centroid. (The point O may be located by the 
method described in Art. 184.) 
From symmetry it is evident 
that the principal axes of the 
central ellipse are XX^ and 
YY\ drawn through O, re- 
spectively parallel and per- 
pendicular to AB. 

With respect to the axis 
YY\ the radius of gyration 
is evidently \ r, the same as 

for the whole circle. Hence two lines parallel to YY' and dis- 
tant ^ r from it arctangents to the central ellipse. 

Again, the radius of gyration of the semicircle with respect 

to AB as an axis is also \ r, the same as for the whole circle. 

To find it for the axis XX\ with Z^ as a center, and radius \ r, 

9 2 '^ 

draw an arc intersectmg XX' at F\ then OF^ = DF —QD'^. 

But DF'i's, equal to the radius of gyration with respect to AB, 

and 0D\^ the distance between XX' and AB \ hence (Art. 196) 

OF is equal to the radius of gyration with respect to XX . 

Hence if two lines are drawn parallel to XX , each at a distance 

from it equal to OF, they will be tangents to the central ellipse. 

The ellipse can now be drawn in the usual manner. 



219. Summary of Results. — By the principles and methods 
developed in the present chapter, inertia-curves can be drawn 
for all the simpler cases that may arise ; namely, whenever a 
pair of conjugate axes can be located by inspection. This will 
be the case whenever the product of inertia can be seen to be 
zero for any pair of axes ; and it includes every case of an area 
possessing an axis of symmetry. 

It is believed that this chapter contains as complete a discus- 
sion as is needed by the student of engineering. Those who 
desire to pursue the subject further may consult other works. 



Fig. 34 

U) 

Scale, 1 incli^Gfect 




PLATE I. 





/ 
DEF 



Scale, 1 inch=6,000 lbs. 
(C) 



Scale: I in. = 4,000 lbs. 





Scale: I in, 



Fig. 42 




PLATE II. 




8.1 



5.8 



4.5 



4.8 



5.7 



o OOOO ooo!) 




Linear Scale, 1 in. = lo ft. 
Force Scale, 1 in.= 80,000 lbs 



Fio:.47 



PLATE III. 



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8.1 



5.8 



4.8 



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4.8 



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i' 



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7.1 



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PLATE IV. 



4.5 



4.5 



7.1 



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o O OOO O O O O I ■■"»'^'-p'-"- 




M 



r 



SS 'Sx^ 




» }l .6 J .d ,V >o fl .n U / 



oo oo o O O o oo oo ^ 



0-f 


i-f 


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87 


e-i 


ri 




6-i 




0-6 


0-f 


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to 


























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X; 


(al 


CT 






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sqi ooo'og-'ni T 'oivos oodoj 

"if OS^'Ul T 'OXDOS JLDOUl'J 

n 




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